1. Definitions
$\newcommand{\kk}{\mathbf{k}}$
First, let me restate your notations and definition.
Fix a Lie algebra $L$ over any commutative ring $\mathbf{k}$. The word "span" shall always mean "$\mathbf{k}$-linear span" from now on. If $U$ is a subset of $L$, then $\kk U$ shall denote the span of $U$.
If $U$ and $V$ are two subsets of $L$, then $\left[U, V\right]_0$ shall mean the set of all Lie brackets $\left[u, v\right]$ with $u \in U$ and $v \in V$. This is a subset of $L$.
If $U$ and $V$ are two $\mathbf{k}$-submodules of $L$, then $\left[U, V\right]$ shall mean the span of the set $\left[U, V\right]_0$. This is a $\mathbf{k}$-submodule of $L$ and contains $\left[U, V\right]_0$ as a subset (but is, in general, greater).
Let $X$ be a subset of $L$. Let me restate the definition of Lie monomials as follows:
Definition 1. We define a sequence $\left(B_1, B_2, B_3, \ldots\right)$ of subsets of $L$ recursively as follows: We set $B_1 = X$; then, for each $n > 1$, we set $B_n = \bigcup\limits_{a + b = n} \left[B_a, B_b\right]_0$, where the $a$ and $b$ in the big union are meant to range over positive integers.
Thus, $B_1 = X$ and $B_2 = \left[B_1, B_1\right]_0 = \left[X, X\right]_0$ and $B_3 = \left[B_1, B_2\right]_0 \cup \left[B_2, B_1\right]_0 = \left[X, \left[X, X\right]_0\right]_0 \cup \left[\left[X, X\right]_0, X\right]_0$ and $B_4 = \left[B_1, B_3\right]_0 \cup \left[B_2, B_2\right]_0 \cup \left[B_3, B_1\right]_0$ and so on.
Definition 2. Set $B = B_1 \cup B_2 \cup B_3 \cup \cdots$. The elements of $B$ are called the Lie monomials in $X$. For each $n \geq 1$, the elements of $B_n$ are called the Lie monomials in $X$ of length $n$.
Next, I shall define what you call "simpler monomials", but I will call them "left-bracketed Lie monomials" instead:
Definition 3. We define a sequence $\left(C_1, C_2, C_3, \ldots\right)$ of subsets of $L$ recursively as follows: We set $C_1 = X$; then, for each $n > 1$, we set $C_n = \left[C_{n-1}, X\right]_0$.
Thus, $C_1 = X$ and $C_2 = \left[C_1, X\right]_0 = \left[X, X\right]_0$ and $C_3 = \left[C_2, X\right]_0 = \left[\left[X, X\right]_0, X\right]_0$ and $C_4 = \left[C_3, X\right]_0 = \left[\left[\left[X, X\right]_0, X\right]_0, X\right]_0$ and so on.
Definition 4. Set $C = C_1 \cup C_2 \cup C_3 \cup \cdots$. The elements of $C$ are called the left-bracketed Lie monomials in $X$.
2. The claim and the proof outline
Now, your claim is the following:
Theorem 1. We have $B \subseteq \kk C$.
The proof is easiest made using the following definition:
Definition 5. We define a sequence $\left(L_1, L_2, L_3, \ldots\right)$ of $\mathbf{k}$-submodules subsets of $L$ recursively as follows: We set $L_1 = \kk X$; then, for each $n > 1$, we set $L_n = \left[L_{n-1}, \kk X\right]$.
Thus, $L_1 = \kk X$ and $L_2 = \left[L_1, \kk X\right] = \left[\kk X, \kk X\right]$ and $L_3 = \left[L_2, \kk X\right] = \left[\left[\kk X, \kk X\right], \kk X\right]$ and so on.
Now, let me split Theorem 1 into the following bite-sized pieces:
Proposition 2. We have $L_n = \kk C_n$ for each $n \geq 1$.
Proposition 3. We have $\left[L_a, L_b\right] \subseteq L_{a+b}$ for any $a \geq 1$ and $b \geq 1$.
Proposition 4. We have $B_n \subseteq L_n$ for each $n \geq 1$.
See below for the detailed proofs of these three propositions as well as the derivation of Theorem 1 from them. But first, here are hints that should suffice if you have any experience with the Lie algebra axioms:
Proposition 2 is proven by straightforward induction on $n$.
To prove Proposition 3, we proceed by induction on $a$. In the induction step, we assume that Proposition 3 is true for $a-1$, and intend to prove it for $a$. It suffices to show that $\left[\left[x, y\right], z\right] \in L_{a+b}$ for all $x \in L_{a-1}$, $y \in X$ and $z \in L_b$ (because $L_a = \left[L_{a-1}, L\right]$ is spanned by elements of the form $\left[x, y\right]$ with $x \in L_{a-1}$ and $y \in X$). But this follows by applying the Jacobi identity
\begin{align}
\left[\left[x, y\right], z\right]
= \left[\left[x, z\right], y\right] - \left[x, \left[z, y\right]\right]
\end{align}
and realizing that both addends $\left[\left[x, z\right], y\right]$ and $- \left[x, \left[z, y\right]\right]$ on the right hand side belong to $L_{a+b}$ (indeed, we have $\left[x, z\right] \in \left[L_{a-1}, L_b\right] \subseteq L_{a+b-1}$ (by the induction hypothesis) and thus $\left[\left[x, z\right], y\right] \in \left[L_{a+b-1}, \kk X\right] = L_{a+b}$ (by the definition of $L_{a+b}$), and we also have $\left[z, y\right] \in \left[L_b, \kk X\right] = L_{b+1}$ (by the definition of $L_{b+1}$) and therefore $\left[x, \left[z, y\right]\right] \in \left[L_{a-1}, L_{b+1}\right] \subseteq L_{a+b}$ (by the induction hypothesis)). So Proposition 3 follows by induction.
Proposition 4 is proven by strong induction on $n$, using Proposition 3.
Combining Proposition 2 with Proposition 4, we obtain $B_n \subseteq L_n = \kk C_n$ for each $n \geq 1$. Thus, $B \subseteq \kk C$, so that Theorem 1 is proven.
3. Formal proofs
Let me prove these Propositions 2, 3 and 4 in detail, just to make sure
everything is exactly as I claimed (you have most likely proven them yourself
by now). This has turned out to be even duller than expected.
We will tacitly use the observation that if $U$ and $V$ are two subsets of
$L$, and if $u\in U$ and $v\in V$, then $\left[ u,v\right] \in\left[
U,V\right] $. (Indeed, if $U$ and $V$ are two subsets of $L$, then the
definition of $\left[ U,V\right] $ shows that $\left[ U,V\right] =\left(
\text{the span of }\left[ U,V\right] _0 \right) =\kk \left( \left[
U,V\right] _0 \right) $. Thus, if $U$ and $V$ are two subsets of $L$, and
if $u\in U$ and $v\in V$, then we have $\left[ u,v\right] \in\left[
U,V\right] _0 \subseteq\kk \left( \left[ U,V\right] _0 \right)
=\left[ U,V\right] $.)
We shall use two simple lemmas:
Lemma 5. Let $S$ be a subset of $L$. Let $M$ be a $\kk$-submodule
of $L$. If $S\subseteq M$, then $\kk S\subseteq M$.
Proof of Lemma 5. Recall that $\kk S$ is the span of $S$, and thus is
the smallest $\kk$-submodule of $L$ that contains $S$ as a subset.
Hence, any $\kk$-submodule of $L$ that contains $S$ as a subset must
contain $\kk S$ as a subset. Applying this to the $\kk$-submodule
$M$, we conclude that $M$ contains $\kk S$ as a subset if $M$ contains
$S$ as a subset. In other words, if $S\subseteq M$, then $\kk S\subseteq
M$. This proves Lemma 5. $\blacksquare$
Lemma 6. Let $U$ and $V$ be two subsets of $L$. Then, $\left[
\kk U,\kk V\right] =\kk \left( \left[ U,V\right]
_0 \right) $.
Proof of Lemma 6. Recall that $\left[ \kk U,\kk V\right] $ is
the span of the subset $\left[ \kk U,\kk V\right] _0 $ (by the
definition of $\left[ \kk U,\kk V\right] $). In other words,
$\left[ \kk U,\kk V\right] =\kk \left( \left[
\kk U,\kk V\right] _0 \right) $. Hence, $\left[ \kk
U,\kk V\right] $ is a $\kk$-submodule of $L$ and satisfies
$\left[ \kk U,\kk V\right] _0 \subseteq\left[ \kk
U,\kk V\right] $.
Let $r\in\left[ \kk U,\kk V\right] _0 $ be arbitrary. We shall
show that $r\in\kk \left( \left[ U,V\right] _0 \right) $.
Indeed, we have $r\in\left[ \kk U,\kk V\right] _0 $; in other
words, $r=\left[ u,v\right] $ for some $u\in\kk U$ and some
$v\in\kk V$ (by the definition of $\left[ \kk U,\kk
V\right] _0 $). Consider these $u$ and $v$.
We have $u\in\kk U$. In other words, $u$ is a $\kk$-linear
combination of the elements of $U$ (by the definition of the span
$\kk U$). In other words, $u=\sum_{i\in I}\lambda_{i}u_{i}$ for some
finite set $I$ and some families $\left( \lambda_{i}\right) _{i\in I}
\in\kk ^{I}$ and $\left( u_{i}\right) _{i\in I}\in U^{I}$. Consider
this $I$ and these families.
We have $v\in\kk U$. In other words, $v$ is a $\kk$-linear
combination of the elements of $V$ (by the definition of the span
$\kk V$). In other words, $v=\sum_{j\in J}\mu_{j}v_{j}$ for some finite
set $J$ and some families $\left( \mu_{j}\right) _{j\in J}\in\kk ^{J}$
and $\left( v_{j}\right) _{j\in J}\in V^{J}$. Consider this $J$ and these families.
Now,
\begin{align*}
r & =\left[ u,v\right] =\left[ \sum_{i\in I}\lambda_{i}u_{i},\sum_{j\in
J}\mu_{j}v_{j}\right] \qquad\left( \text{since }u=\sum_{i\in I}\lambda
_{i}u_{i}\text{ and }v=\sum_{j\in J}\mu_{j}v_{j}\right) \\
& =\sum_{i\in I}\lambda_{i}\sum_{j\in J}\mu_{j}\left[ u_{i},v_{j}\right]
\qquad\left( \text{since the Lie bracket is }\kk \text{-bilinear}
\right) \\
& =\sum_{\left( i,j\right) \in I\times J}\lambda_{i}\mu_{j}
\underbrace{\left[ u_{i},v_{j}\right] }_{\substack{\in\left[ U,V\right]
_0 \\\text{(since }u_{i}\in U\text{ and }v_{j}\in V\text{)}}}\in\sum_{\left(
i,j\right) \in I\times J}\lambda_{i}\mu_{j}\left[ U,V\right] _0 .
\end{align*}
Thus, $r$ is a $\kk$-linear combination of elements of $\left[
U,V\right] _0 $. In other words, $r$ belongs to the span of $\left[
U,V\right] _0 $. In other words, $r\in\kk \left( \left[ U,V\right]
_0 \right) $ (because $\kk \left( \left[ U,V\right] _0 \right) $
is the span of $\left[ U,V\right] _0 $).
Now, forget that we fixed $r$. We thus have proven that $r\in\kk \left(
\left[ U,V\right] _0 \right) $ for each $r\in\left[ \kk
U,\kk V\right] _0 $. In other words, $\left[ \kk
U,\kk V\right] _0 \subseteq\kk \left( \left[ U,V\right]
_0 \right) $. Thus, Lemma 5 (applied to $M=\kk \left( \left[
U,V\right] _0 \right) $ and $S=\left[ \kk U,\kk V\right]
_0 $) yields that $\kk \left( \left[ \kk U,\kk V\right]
_0 \right) \subseteq\kk \left( \left[ U,V\right] _0 \right) $.
Now, recall that $\left[ \kk U,\kk V\right] =\kk \left(
\left[ \kk U,\kk V\right] _0 \right) \subseteq\kk
\left( \left[ U,V\right] _0 \right) $.
On the other hand, let $q\in\left[ U,V\right] _0 $. We shall show that
$q\in\left[ \kk U,\kk V\right] $.
In fact, we have $q\in\left[ U,V\right] _0 $; in other words, $q=\left[
\widetilde{u},\widetilde{v}\right] $ for some $\widetilde{u}\in U$ and some
$\widetilde{v}\in V$ (by the definition of $\left[ U,V\right] _0 $).
Consider these $\widetilde{u}$ and $\widetilde{v}$. From $\widetilde{u}\in
U\subseteq\kk U$ and $\widetilde{v}\in V\subseteq\kk V$, we obtain
$\left[ \widetilde{u},\widetilde{v}\right] \in\left[ \kk
U,\kk V\right] $. Thus, $q=\left[ \widetilde{u},\widetilde{v}\right]
\in\left[ \kk U,\kk V\right] $.
Forget that we fixed $q$. We thus have proven that $q\in\left[ \kk
U,\kk V\right] $ for each $q\in\left[ U,V\right] _0 $. In other
words, $\left[ U,V\right] _0 \subseteq\left[ \kk U,\kk
V\right] $. Thus, Lemma 5 (applied to $M=\left[ \kk U,\kk
V\right] $ and $S=\left[ U,V\right] _0 $) yields that $\kk \left(
\left[ U,V\right] _0 \right) \subseteq\left[ \kk U,\kk
V\right] $. Combining this with $\left[ \kk U,\kk V\right]
\subseteq\kk \left( \left[ U,V\right] _0 \right) $, we obtain
$\left[ \kk U,\kk V\right] =\kk \left( \left[
U,V\right] _0 \right) $. This proves Lemma 6. $\blacksquare$
Proof of Proposition 2. We shall prove Proposition 2 by induction on $n$:
Induction base: The definition of $C_{1}$ yields $C_{1}=X$. The definition
of $L_{1}$ yields $L_{1}=\kk \underbrace{X}_{=C_{1}}=\kk C_{1}$.
In other words, Proposition 2 holds for $n=1$. This completes the induction base.
Induction step: Fix a positive integer $m>1$. Assume that Proposition 2
holds for $n=m-1$. We must prove that Proposition 2 holds for $n=m$.
We have assumed that Proposition 2 holds for $n=m-1$. In other words,
$L_{m-1}=\kk C_{m-1}$. The recursive definition of the $C_{n}$ yields
$C_{m}=\left[ C_{m-1},X\right] _0 $. Hence, $\kk C_{m}=\kk
\left( \left[ C_{m-1},X\right] _0 \right) $. But Lemma 6 (applied to
$U=C_{m-1}$ and $V=X$) yields $\left[ \kk C_{m-1},\kk X\right]
=\kk \left( \left[ C_{m-1},X\right] _0 \right) $. Comparing these
two equalities, we obtain $\kk C_{m}=\left[ \kk C_{m-1}
,\kk X\right] $.
But the recursive definition of the $L_{n}$ yields $L_{m}=\left[
\underbrace{L_{m-1}}_{=\kk C_{m-1}},\kk X\right] =\left[
\kk C_{m-1},\kk X\right] $. Comparing these two equalities, we
obtain $L_{m}=\kk C_{m}$. In other words, Proposition 2 holds for $n=m$.
This completes the induction step. Thus, Proposition 2 is proven.
$\blacksquare$
Proof of Proposition 3. We shall prove Proposition 3 by induction on $a$:
Induction base: Let $b\geq1$ be an integer. We shall prove that $\left[
L_{1},L_{b}\right] \subseteq L_{1+b}$.
Indeed, the recursive definition of the $L_{n}$ yields $L_{b+1}=\left[
L_{b},\kk X\right] $ and $L_{1}=\kk X$.
Let $r\in\left[ L_{1},L_{b}\right] _0 $ be arbitrary. We shall show that
$r\in L_{b+1}$.
Indeed, we have $r\in\left[ L_{1},L_{b}\right] _0 $; in other words,
$r=\left[ u,v\right] $ for some $u\in L_{1}$ and some $v\in L_{b}$ (by the
definition of $\left[ L_{1},L_{b}\right] _0 $). Consider these $u$ and $v$.
We have $\left[ \underbrace{v}_{\in L_{b}},\underbrace{u}_{\in L_{1}}\right]
\in\left[ L_{b},\underbrace{L_{1}}_{=\kk X}\right] =\left[
L_{b},\kk X\right] =L_{b+1}$ (since $L_{b+1}=\left[ L_{b}
,\kk X\right] $). But $r=\left[ u,v\right] =-\left[ v,u\right] $
(since the Lie bracket is antisymmetric). Hence, $r=-\underbrace{\left[
v,u\right] }_{\in L_{b+1}}\in-L_{b+1}\subseteq L_{b+1}$ (since $L_{b+1}$ is a
$\kk$-submodule of $L$).
Now, forget that we fixed $r$. We thus have proven that $r\in L_{b+1}$ for
each $r\in\left[ L_{1},L_{b}\right] _0 $. In other words, $\left[
L_{1},L_{b}\right] _0 \subseteq L_{b+1}$. Hence, Lemma 5 (applied to
$M=L_{b+1}$ and $S=\left[ L_{1},L_{b}\right] _0 $) yields that
$\kk \left( \left[ L_{1},L_{b}\right] _0 \right) \subseteq L_{b+1}
$. But $\left[ L_{1},L_{b}\right] $ is the span of $\left[ L_{1}
,L_{b}\right] _0 $ (by the definition of $\left[ L_{1},L_{b}\right] $); in
other words, $\left[ L_{1},L_{b}\right] =\kk \left( \left[
L_{1},L_{b}\right] _0 \right) $. Hence, $\left[ L_{1},L_{b}\right]
=\kk \left( \left[ L_{1},L_{b}\right] _0 \right) \subseteq
L_{b+1}=L_{1+b}$.
Now, forget that we fixed $b$. We thus have proven that $\left[ L_{1}
,L_{b}\right] \subseteq L_{1+b}$ for all $b\geq1$. In other words,
Proposition 3 holds for $a=1$. This completes the induction base.
Induction step: Fix an integer $c>1$. Assume that Proposition 3 holds for
$a=c-1$. We must prove that Proposition 3 holds for $a=c$.
Let $b\geq1$ be an integer. We shall show that $\left[ L_{c},L_{b}\right]
\subseteq L_{c+b}$.
Proposition 2 (applied to $n=b$) yields $L_{b}=\kk C_{b}$. Proposition 2
(applied to $n=c$) yields $L_{c}=\kk C_{c}$. Now,
\begin{align*}
\left[ \underbrace{L_{c}}_{=\kk C_{c}},\underbrace{L_{b}}
_{=\kk C_{b}}\right] =\left[ \kk C_{c},\kk C_{b}\right]
=\kk \left( \left[ C_{c},C_{b}\right] _0 \right)
\end{align*}
(by Lemma 6, applied to $U=C_{c}$ and $V=C_{b}$).
Now, let $r\in\left[ C_{c},C_{b}\right] _0 $ be arbitrary. We shall show
that $r\in C_{c+b}$.
We have $r\in\left[ C_{c},C_{b}\right] _0 $. In other words, $r=\left[
u,z\right] $ for some $u\in C_{c}$ and some $z\in C_{b}$ (by the definition
of $\left[ C_{c},C_{b}\right] _0 $). Consider these $u$ and $z$.
We have $u\in C_{c}=\left[ C_{c-1},X\right] _0 $ (by the recursive
definition of the $C_{n}$, since $c>1$). In other words, $u=\left[
x,y\right] $ for some $x\in C_{c-1}$ and some $y\in X$ (by the definition of
$\left[ C_{c-1},X\right] _0 $). Consider these $x$ and $y$.
The recursive definition of the $L_{n}$ yields $L_{c+b}=\left[ L_{c+b-1}
,\kk X\right] $ and $L_{b+1}=\left[ L_{b},\kk X\right] $.
We can apply Proposition 3 to $c-1$ instead of $a$ (since we have assumed that
Proposition 3 holds for $a=c-1$). We thus obtain $\left[ L_{c-1}
,L_{b}\right] \subseteq L_{\left( c-1\right) +b}=L_{c+b-1}$.
We can apply Proposition 3 to $c-1$ and $b+1$ instead of $a$ and $b$ (since we
have assumed that Proposition 3 holds for $a=c-1$). We thus obtain $\left[
L_{c-1},L_{b+1}\right] \subseteq L_{\left( c-1\right) +\left( b+1\right)
}=L_{c+b}$.
Proposition 2 (applied to $n=c-1$) yields $L_{c-1}=\kk C_{c-1}$. From
$x\in C_{c-1}\subseteq\kk C_{c-1}=L_{c-1}$ and $z\in C_{b}
\subseteq\kk C_{b}=L_{b}$, we obtain $\left[ \underbrace{x}_{\in
L_{c-1}},\underbrace{z}_{\in L_{b}}\right] \in\left[ L_{c-1},L_{b}\right]
\subseteq L_{c+b-1}$. Combining this with $y\in X\subseteq\kk X$, we
obtain $\left[ \underbrace{\left[ x,z\right] }_{\in L_{c+b-1}
},\underbrace{y}_{\in\kk X}\right] \in\left[ L_{c+b-1},\kk
X\right] =L_{c+b}$.
From $z\in L_{b}$ and $y\in X\subseteq\kk X$, we obtain $\left[
\underbrace{z}_{\in L_{b}},\underbrace{y}_{\in\kk X}\right] \in\left[
L_{b},\kk X\right] =L_{b+1}$. Hence, $\left[ \underbrace{x}_{\in
L_{c-1}},\underbrace{\left[ z,y\right] }_{\in L_{b+1}}\right] \in\left[
L_{c-1},L_{b+1}\right] \subseteq L_{c+b}$.
But recall that
\begin{align*}
r & =\left[ \underbrace{u}_{=\left[ x,y\right] },z\right] =\left[
\left[ x,y\right] ,z\right] =\underbrace{\left[ \left[ x,z\right]
,y\right] }_{\in L_{c+b}}-\underbrace{\left[ x,\left[ z,y\right] \right]
}_{\in L_{c+b}}\\
& \qquad\qquad\left(
\begin{array}
[c]{c}
\text{since the Jacobi identity yields}\\
\left[ \left[ x,z\right] ,y\right] =\left[ \left[ x,y\right] ,z\right]
+\left[ x,\left[ z,y\right] \right]
\end{array}
\right) \\
& \in L_{c+b}-L_{c+b}\subseteq L_{c+b}\qquad\left( \text{since }
L_{c+b}\text{ is a }\kk \text{-submodule of }L\right) .
\end{align*}
Now, forget that we fixed $r$. We thus have shown that $r\in L_{c+b}$ for each
$r\in\left[ C_{c},C_{b}\right] _0 $. In other words, $\left[ C_{c}
,C_{b}\right] _0 \subseteq L_{c+b}$.
Hence, Lemma 5 (applied to $M=L_{c+b}$ and $S=\left[ C_{c},C_{b}\right]
_0 $) yields that $\kk \left( \left[ C_{c},C_{b}\right] _0 \right)
\subseteq L_{c+b}$. Now, recall that $\left[ L_{c},L_{b}\right]
=\kk \left( \left[ C_{c},C_{b}\right] _0 \right) \subseteq L_{c+b}$.
Now, forget that we fixed $b$. We thus have proven that $\left[ L_{c}
,L_{b}\right] \subseteq L_{c+b}$ for all $b\geq1$. In other words,
Proposition 3 holds for $a=c$. This completes the induction step. Hence,
Proposition 3 is proven by induction. $\blacksquare$
Proof of Proposition 4. We shall prove Proposition 4 by strong induction on
$n$. Thus, we fix an integer $m\geq1$. We assume that Proposition 4 holds for
all $n<m$. We want to prove that Proposition 4 holds for $n=m$ as well.
Let $r\in B_{m}$. We shall prove that $r\in L_{m}$.
If $m=1$, then this holds for easy reasons (in fact, if $m=1$, then
$B_{m}=B_{1}=X\subseteq\kk X=L_{1}=L_{m}$ (since $1=m$), and therefore
$r\in B_{m}\subseteq L_{m}$). Thus, for the rest of this proof, we WLOG assume
that we don't have $m=1$. Hence, $m>1$.
Thus, the recursive definition of $B_{m}$ yields $B_{m}=\bigcup\limits_{a+b=m}
\left[ B_{a},B_{b}\right] _0 $. Hence, $r\in B_{m}=\bigcup\limits_{a+b=m}
\left[ B_{a},B_{b}\right] _0 $. In other words, $r\in\left[ B_{a}
,B_{b}\right] _0 $ for some pair $\left( a,b\right) $ of positive integers
satisfying $a+b=m$. Consider this pair $\left( a,b\right) $. We have $a>0$
(since $a$ is a positive integer) and thus $a+b>b$, so that $b<a+b=m$.
Similarly, $a<m$.
We have assumed that Proposition 4 holds for all $n<m$. Hence, we can apply
Proposition 4 to $n=a$ (since $a<m$). We thus obtain $B_{a}\subseteq L_{a}$.
The same argument (applied to $b$ instead of $a$) yields $B_{b}\subseteq
L_{b}$.
But $r\in\left[ B_{a},B_{b}\right] _0 $. In other words, $r=\left[
u,v\right] $ for some $u\in B_{a}$ and some $v\in B_{b}$ (by the definition
of $\left[ B_{a},B_{b}\right] _0 $). Consider these $u$ and $v$. We have
$u\in B_{a}\subseteq L_{a}$ and $v\in B_{b}\subseteq L_{b}$. Now, $r=\left[
\underbrace{u}_{\in L_{a}},\underbrace{v}_{\in L_{b}}\right] \in\left[
L_{a},L_{b}\right] \subseteq L_{a+b}$ (by Proposition 3). In view of $a+b=m$,
this rewrites as $r\in L_{m}$.
Now, forget that we fixed $r$. We thus have proven that $r\in L_{m}$ for each
$r\in B_{m}$. In other words, $B_{m}\subseteq L_{m}$. In other words,
Proposition 4 holds for $n=m$. This completes the induction proof. Thus,
Proposition 4 is proven by strong induction. $\blacksquare$
Proof of Theorem 1. Let $r\in B$. Thus, $r\in B=B_{1}\cup B_{2}\cup
B_{3}\cup\cdots$. In other words, $r\in B_{n}$ for some integer $n\geq1$. Fix
this $n$.
Now, $r\in B_{n}\subseteq L_{n}$ (by Proposition 4), so that $r\in
L_{n}=\kk C_{n}$ (by Proposition 2). But $C_{n}\subseteq C_{1}\cup
C_{2}\cup C_{3}\cup\cdots=C$ (since $C=C_{1}\cup C_{2}\cup C_{3}\cup\cdots$),
so that $r\in\kk \underbrace{C_{n}}_{\subseteq C}\subseteq\kk C$.
Now, forget that we fixed $r$. We thus have proven that $r\in\kk C$ for
each $r\in B$. In other words, $B\subseteq\kk C$. This proves Theorem 1.
$\blacksquare$
