Summary of solution : the originality of your question consists in its use of
the floor function to round, when most of the classical results on continued fractions use
the nearest integer function. The idea is to reduce the former to the latter.
Details :
Let $M_n,a_n$ be as in the OP. Our goal is to show that
$$
M_n=a_n \tag{1}
$$
As a convenience, let us use the "normalized" number $\alpha=\sqrt{2}-1$ (which is in the unit interval) rather than $\sqrt{2}$ itself.
It will suffice to show :
$$
\lbrace q\alpha \rbrace \geq \lbrace a_n\alpha \rbrace
\textrm{ for } 0 \lt q \lt a_{n+1} \tag{2}
$$
We will use a sequence closely related to the continued fraction expansion of $\alpha$, namely $g_n=\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}$. Note that $(g_n)$ satisfies
$g_0=0,g_1=1$ and $g_{n+2}-2g_{n+1}+g_n=0$, so its values (after $g_1$) are positive integers. The following properties are easy to show by a direct computation or by induction :
$(i)$ $g_ng_{n+2}-g_{n+1}^2=(-1)^{n+1}$
$(ii)$ $a_n=g_n$ when $n$ is odd
$(iii)$ $a_n=g_{n+1}-g_n$ when $n$ is even
$(iv)$ $\frac{g_n-g_{n+1}\alpha}{g_{n+1}-g_{n+2}\alpha}=1-\sqrt{2}$
$(v)$ $\frac{a_{n+2}\alpha -(g_{n+2}-g_{n+1})}{a_n\alpha -(g_n-g_{n-1})}=3-2\sqrt{2}$,
when $n$ is even.
The following lemma is a very classic continued fraction inequality :
LEMMA. Let $p,q$ be integers with $0 \lt q \lt g_{n+2}$. Then $|p-q\alpha| \geq |g_n-g_{n+1}\alpha|$. Further, if $(p,q)\neq (g_n,g_{n+1})$, the inequality can be strenghtened to $|p-q\alpha| \geq |g_n-g_{n+1}\alpha|+|g_{n+1}-g_{n+2}\alpha|$.
Proof of lemma. By (i), the matrix $\left(\begin{matrix} g_n & g_{n+1} \\ g_{n+1} & g_{n+2} \end{matrix}\right)$ is unimodular. It follows that there are integers $u,v$ such that $p=g_n u+g_{n+1}v$ and $q=g_{n+1}u+g_{n+2}v$. I claim that $uv\leq 0$ ; otherwise, $g_{n+1}u$ and $g_{n+2}v$ would both be nonzero and have the same sign, forcing $|q| = |g_{n+1}u|+|g_{n+2}v| \geq |g_{n+2}|$ which is impossible. So $uv\leq 0$, and also
$$
p-q\alpha= u(g_n-g_{n+1}\alpha) + v (g_{n+1}-g_{n+2}\alpha) \tag{3}
$$
The numbers $u(g_n-g_{n+1}\alpha)$ and $v(g_{n+1}-g_{n+2}\alpha)$ have the same sign, so
$$
|p-q\alpha|= |u(g_n-g_{n+1}\alpha)| + |v (g_{n+1}-g_{n+2}\alpha)| \tag{4}
$$
Note that $u\neq 0$ because of $q=g_{n+1}u+g_{n+2}v$ and $|q|<g_{n+2}$. So $|u|\geq 1$ and the first inequality follows. If equality holds, we must have $|u|=1$ and $v=0$, from which we easily deduce $(p,q)=(g_n,g_{n+1})$. Suppose now that $(p,q)\neq (g_n,g_{n+1})$. If $|u|\geq 2$, then from (4) we deduce $|p-q\alpha| \geq 2|g_n-g_{n+1}\alpha|$. By (iv), the second inequality holds in this case.
Otherwise $u=1$ and from the reasoning above we have $v\neq 0$, so $|v|\geq 1$ and from (4) again we deduce $|p-q\alpha| \geq |g_n-g_{n+1}\alpha|+|g_{n+1}-g_{n+2}\alpha|$, which finishes the proof of the lemma.
Let us now deduce (2) from the lemma. Suppose first that $n$ is odd. By $(ii)$, $\lbrace a_n\alpha \rbrace=\lbrace g_n\alpha \rbrace=g_{n+1}\alpha-g_n$ by lemma, (iv) and the fact that $g_{n+1}\alpha-g_n >0$.
The lemma also shows that $\lbrace q \alpha \rbrace \geq \lbrace a_n \alpha \rbrace$
for $0 \lt q \lt g_{n+2}=a_{n+2}$, as wished.
Suppose now that $n$ is even. Using the lemma with $n-1$ in place of $n$, we have that for any $0 \lt q \lt g_{n+1}=a_{n+1}$ and $(p,q)\neq (g_{n-1},g_n)$, then
$$\begin{array}{lcl}
|p-q\alpha| &\geq& |g_{n-1}-g_{n}\alpha|+|g_{n}-g_{n+1}\alpha| \\
&=& g_{n-1}-g_{n}\alpha+g_{n+1}\alpha-g_n \\
&=& a_n\alpha -(g_n-g_{n-1}) \\
&=& \lbrace a_n \alpha \rbrace \textrm{ by } (v)
\end{array}$$
This finishes the proof.
