I will show you that if $X$ is Banach and Y is a closed subspace then $X/Y$ is Banach. From this hopefully you can extrapolate to your case. Now, you just need to define an inner product on your factor space.
To that end, let $X$ be Banach and $Y$ a closed subspace, and let $\{X_n\} \subset X/Y$ be an absolutely convergent series in $X/Y$, i.e, $\sum_{n \in \mathbb{N}}{\|X_n\|_{X/Y}} < \infty$. Recall that our quotient norm is defined as
$$\|x\|_{X/Y}=\inf_{u \in x+Y}{\|u\|_X}=\inf_{y \in Y}{\|x+y\|_X}.$$
Then by definition of the quotient norm for any $X_n \in X/Y$ there exists $x_n \in X_n$ such that
$$\|x_n\|_{X} \leq \|X_n\|_{X/Y}+2^{-n}.$$
Then we have that $\sum_{n \in \mathbb{N}}{\|x_n\|_{X}} < \infty$ and hence is absolutely convergent and therefore $\exists x \in X$ such that $x=\sum_{n \in \mathbb{N}}{x_n}$. Note this follows from the fact that $X$ is a Banach space.
Now, define $X=x+Y$ and $S_k=\sum_{n=1}^{k}{x_n}+Y$. Now, note that
$$\|U-V\|_{X/Y}=\inf_{k \in (u-v)+Y}{\|k\|_{X}} = \inf_{y \in Y}{\|(u-v)+y\|_{X}} \leq \inf_{y \in Y}{ \|u-v\|+\|y\|}$$
and lastly, we note that $y=0 \in Y$ and hence we have the right most term is equal to $\|u-v\|_{X}$. Hence, we get that
$$\|X-S_k\|_{X/Y} \leq \|x-s_k\|_{X} \underset{k \to \infty}{\longrightarrow} 0,$$
where $s_k=\sum_{n=1}^{k}{x_n}$. Hence, we have that our series in our factor space is convergent and thus $X/Y$ is Banach
Per my comment below:
You can also do this by noting that Given a closed subspace $Y \subset X$,
$X/Y \simeq Y^{\perp}$ and the result follows trivially.