A quotient of a Banach space by a closed subspace is Banach

Since $({\xi }_n)$ is Cauchy in $X\mathrm{/}M$, choose a strictly increasing sequence $(n_k)$ such that $\mathrm{\parallel }{\xi }_{n_{k+1}}-{\xi }_{n_k}{\mathrm{\parallel }}_{X\mathrm{/}M}<2^{-k}$ for every $k\ge 0$.

For each $k$, choose $u_k\in X$ representing ${\xi }_{n_{k+1}}-{\xi }_{n_k}$ and satisfying $\mathrm{\parallel }{u}_k\mathrm{\parallel }<2^{-k}+2^{-2k}$. This is possible by [L0], [L2], and step 1.1.

The series ${\sum }_k{u}_k$ is absolutely convergent because ${\sum }_k(2^{-k}+2^{-2k})$ converges. Since $X$ is Banach, [L4] gives a vector $u\in X$ with ${\sum }_k{u}_k=u$.

Let $s_j:={\sum }_{k=0}^{j-1}{u}_k$. Because each $u_k$ represents ${\xi }_{n_{k+1}}-{\xi }_{n_k}$, the coset $q(s_j)$ equals ${\xi }_{n_j}-{\xi }_{n_0}$. Therefore ${\xi }_{n_j}={\xi }_{n_0}+q(s_j)$. Since $s_j\to u$ in $X$, the tails satisfy $\mathrm{\parallel }u-s_j\mathrm{\parallel }\to 0$, and [L2] gives $\mathrm{\parallel }q(u)-q(s_j){\mathrm{\parallel }}_{X\mathrm{/}M}\le \mathrm{\parallel }u-s_j\mathrm{\parallel }$. Hence ${\xi }_{n_j}\to \eta :={\xi }_{n_0}+q(u)$ in $X\mathrm{/}M$.

The whole sequence $({\xi }_n)$ converges to $\eta$. Given $\epsilon >0$, choose $J$ so that $\mathrm{\parallel }{\xi }_n-{\xi }_m{\mathrm{\parallel }}_{X\mathrm{/}M}<\epsilon \mathrm{/}2$ for all $m,n\ge J$, and also choose $k$ with $n_k\ge J$ and $\mathrm{\parallel }{\xi }_{n_k}-\eta {\mathrm{\parallel }}_{X\mathrm{/}M}<\epsilon \mathrm{/}2$ from step 4.1. Then for every $n\ge J$,

Every Cauchy sequence in $X\mathrm{/}M$ converges, so $X\mathrm{/}M$ is Banach by [L1].