(I'm adding this answer since it shows that the statement as written is not completely correct, because it shows a different style of proof, and because I discovered a direct connection with induction.)
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
$Let's first restate the statement in a slightly different notation, with $\;B_n\;$ expanded, and implicitly assuming that $\;k,n\;$ are integers with $\;k \ge 1\;$ and $\;n \ge 2\;$:
$$
\tag 0
\langle \cup n :: A_n \rangle
\;=\;
\langle \cup n :: A_n - \langle \cup k : k < n : A_k \rangle \rangle
$$
The most straightforward approach for proofs like this is to start with the most complex side, and calculate the elements $\;x\;$ of that set:
$$\calc
x \in \langle \cup n :: A_n - \langle \cup k : k < n : A_k \rangle \rangle
\calcop\equiv{expand definition of $\;\cup\;$ twice, and of $\;-\;$}
\langle \exists n :: x \in A_n \land \lnot \langle \exists k : k < n : x \in A_k \rangle \rangle
\calcop\equiv{logic: rewrite using DeMorgan twice, and 'shunting'}
\lnot \langle \forall n : \langle \forall k : k < n : x \not\in A_k \rangle : x \not\in A_n \rangle
\tag{*}
\endcalc$$
The last step was inspired by the shape of the formula, since the last statement reminds us of the principle of complete induction (see Wikipedia and another question):
$$
\tag 1
\langle \forall i : i \in V \land \langle \forall j : j \in V \land j < i : P_j \rangle : P_i \rangle
\;\equiv\;
\langle \forall i : i \in V : P_i \rangle
$$
where $\;i,j\;$ range over the same set $\;V\;$ with an ordering $\;<\;$. (Often $\ref 1$ is presented only with the $\;\Rightarrow\;$ direction, but since the other direction is trivial, it is usually simpler to use the stronger more symmetrical version.)
However, there is a difference between $\ref *$ and $\ref 1$: in $\ref *$ the dummies do not range over the same set, since $\;n \in \mathbb N_{\ge 2}\;$ but $\;k \in \mathbb N_{\ge 1}\;$. So we cannot use $\ref 1$ directly.
Therefore, let's try to rewrite $\ref *$ so that we can use $\ref 1$:
$$\calc
\tag{*}
\lnot \langle \forall n : \langle \forall k : k < n : x \not\in A_k \rangle : x \not\in A_n \rangle
\calcop\equiv{logic: split off $\;k = 1\;$}
\lnot \langle \forall n : x \not\in A_1 \land \langle \forall k : 2 \le k \land k < n : x \not\in A_k \rangle : x \not\in A_n \rangle
\calcop\equiv{logic: DeMorgan; move $\;x \not\in A_1\;$ out of $\;\exists n\;$; DeMorgan}
x \not\in A_1 \land \lnot \langle \forall n : \langle \forall k : 2 \le k \land k < n : x \not\in A_k \rangle : x \not\in A_n \rangle
\calcop{\tag{**} \equiv}{using induction principle $\ref 1$ on $\;\mathbb N_{\ge 2}\;$ with $\;P_n := x \not\in A_n\;$}
x \not\in A_1 \land \lnot \langle \forall n :: x \not\in A_n \rangle
\calcop\equiv{logic: DeMorgan; definitions of $\;\cup\;$ and $\;-\;$}
x \in \langle \cup n :: A_n \rangle - A_1
\endcalc$$
By set extensionality, the above calculation proves
$$
\tag 2
\langle \cup n :: A_n \rangle - A_1
\;=\;
\langle \cup n :: A_n - \langle \cup k : k < n : A_k \rangle \rangle
$$
and the key step was $\ref{**}$ where we applied the principle of complete induction.
Finally, this implies that $\ref 0$ is not true: now that we know $\ref 2$ it is easy to find a counterexample, e.g., if $\;A_1\;$ and $\;A_2\;$ share an element, then that element is in the left hand side of $\ref 0$ but not in its right hand side.
