I want to prove the following remark (found in https://www.uio.no/studier/emner/matnat/math/MAT4400/v19/pensumliste/ela-190523.pdf)
My attempt:
Since $T$ is a compact operator, $T(B)$ is precompact for every bounded set $B \subseteq X$. Thus, $\overline{T(B)}$ is compact, where $\overline{T(B)}$ denotes the closure of $T(B)$.
Since $T(B) \subseteq T(X)$, $\overline{T(B)} \subseteq \overline{T(X)}$.
Thus, $\overline{T(X)}$ has a compact subset, namely $\overline{T(B)}$.
Since compact sets of metric spaces are always separable, $\overline{T(B)}$ has a countable dense subset $A$.
But then $A \subseteq \overline{T(X)}$ is a countable dense subset, hence $\overline{T(X)}$ is separable.
However, I don't think I used that $H$ is a Hilbert space anywhere, which leads me to believe my solution is wrong.