I have this problem:
Let $H$ be a separable Hilbert space and $K$ a subset. Show that $K$ is compact if and only if the following three conditions are met: $K$ is closed; $K$ is bounded; and, for all succession $x_n \in K$, if $x_n$ weakly converges, then converges in norm.
I did the proof and I think it's fine. My problem is that I don´t know what changes if H is not separable, since at no time did I use it in the proof.
This result holds regardless of whether $H$ is separable. Indeed, the general case actually follows easily from the separable case. For if $K$ is not compact, this is witnessed by a countable subset of $K$, namely a sequence $(x_n)$ in $K$ with no convergent subsequence. The closed span $H_0$ of $\{x_n\}$ is then a separable subspace of $H$ such that $K\cap H_0$ is not compact. It follows that $K\cap H_0$ is not closed, not bounded, or has a weakly convergent sequence that does not converge in norm, and thus the same is true of $K$. The converse is similar: if $K$ is not closed, not bounded, or has a weakly convergent sequence that does not converge in norm, then this is witnessed by a countable subset of $K$ and so you get a separable subspace $H_0$ such that $H_0\cap K$ is not compact and so $K$ cannot be compact.