This is a theorem from Megginson's An Introduction to Banach Space Theory. I'll cut right to the chase:
All that needs to be checked is that $X$ is complete if both $M$ and $X/M$ are. (and by $X/M$ the quotient space is meant.)
Here's the proof:
Suppose that $M$ and $X/M$ are Banach spaces. Let $(x_n)$ be a Cauchy sequence in $X$. Since $\lVert (x_n- X_m) +M \rVert \leq \lVert x_n- x_m \rVert $ whenever $m,n \in N$, the sequence $( X_n + M)$ is Cauchy in $X/ M$ and so converges to some $y + M$. By Proposition 1.7.6 (b), there is for each positive integer $n$ some $y_n$ in $X$ such that $y_n +M = (x_n- y) +M$ and $\lVert y_n \rVert < \lVert (x_n -y) +M \rVert + 2^{-n}$. Then $\lim_n y_n = 0$, so $(x_n- y_n- y)$ is a Cauchy sequence in $M$ and therefore has a limit $z$ in $M$. It follows that $X_n = (x_n- y_n- y) + y_n + y \rightarrow z + y$ as $n \rightarrow \infty$, so $X$ is complete. $\quad \square$
Question 1: Why $\lim_n y_n = 0$? Perhaps because $\sum_n \lVert y_n \rVert < \infty$ for some secret reason?
Question 2: Why is $(x_n- y_n- y)$ a Cauchy sequence in $M$?
Thanks.
1) Since $x_{n}+M\rightarrow y+M$, then for large $n$, $\|x_{n}-y+M\|<\epsilon$, and we can pick also that $2^{-n}<\epsilon$, then $\|y_{n}\|<2\epsilon$ for all such $n$.
2) $\|(x_{n}-y_{n}-y)-(x_{m}-y_{m}-y)\|\leq\|x_{n}-x_{m}\|+\|y_{n}-y_{m}\|\rightarrow 0$ as $n,m\rightarrow\infty$ because it has been noted that both $\{x_{n}\}$ and $\{y_{n}\}$ are Cauchy.