The space of Hilbert-Schmidt operators form a Banach space I am having trouble proving the following result:

Show that the space $X$ of bounded operators on a separable Hilbert
space into itself for which the Hilbert-Schmidt norm is finite, is a
Banach space

My attempt:
Let $H$ be a separable Hilbert space. Take a Cauchy sequence $\{A_n\}_{n \ge 1}$ of such operators (bounded + Hilbert-Schmidt). Note that $A_n : H \to H$ for each $n$. Being Cauchy means: $$ \forall \epsilon > 0, \exists N \in \mathbb{N} : n,m > N \implies \|A_{n} - A_{m}\| < \epsilon $$ where $\|\cdot\|$ is the operator norm. My idea is to define the limit operator as: $$ A(u) := \lim_{n \to \infty} A_{n}(u) \text{ for each $u \in H$} $$ Then we need to show that $\|A_n - A\| \to 0$ and $A$ is bounded (i.e. $\|A\| < \infty$). This is sufficent due to the inequality $\|A\| \le \|A\|_{H-S}$ .
But I'm not sure how to prove $\|A_n - A\| \to 0$. I also think there's a better way to choose the candidate limit. Can anyone guide me on how to prove the result?
Thank you.
 A: This is not true for the operator norm. A norm limit (in the operator norm) of Hilbert-Schmidt operators is not necessarily Hilbert-Schmidt; already norm limits of finite-rank operators give you any compact operator. That is, the operator-norm closure of the space of Hilbert-Schmidt operators is $K(H)$, the compact operators.
But that's not what you are being asked. You are asked to show that the Hilbert-Schmidt operators, with the Hilbert-Schmidt norm, form a Banach space.
By choosing a subsequence if necessary, you may assume that $\|A_{n+1}-A_m\|<2^{-n}$. Pointwise, you have
$$\tag1
A=A_1+\sum_n(A_{n+1}-A_n).
$$
This converges in the Hilbert-Schmidt norm because $\sum_n\|A_{n+1}-A_n\|_{\rm HS}\leq\sum_n 2^{-n}=1$. Which in particular shows that $A$ is Hilbert-Schmidt. And we have
$$
\|A-A_n\|_{\rm HS}=\Big\|\sum_{k\geq n}A_{k+1}-A_k\Big\|_{\rm HS}\leq\sum_{k\geq n}\|A_{k+1}-A_k\|_{\rm HS}\leq\sum_{k\geq n}2^{-k}=2^{-n+1}\to0. 
$$
So $A$ is the Hilbert-Schmidt norm limit of a sequence of $\{A_n\}$, and as $\{A_n\}$ is Cauchy it has to converge to $A$.
