Subsets of separable Hilbert spaces are separable

Question

Let Y be a subspace of a Hilbert space, H. Then, if H is separable then so is Y. More generally, every subset of a separable inner product space is separable.

I have been trying to prove the above result. My attempt at the proof is as follows:

Since $H$ is separable, there exists a countable subset, $W$ of $H$ such that $\overline{W} = H$.

Let $W'=W\cap Y$. Since W is a countable set then $W'=W\cap Y$ is also a countable set. Clearly, $W' \subset Y$

Now to show: $\overline{W'} = Y$

$W'=W\cap Y \implies \overline{W'} = \overline{W \cap Y} = \overline{W}\cup\overline{Y} = H \cup \overline{Y}$

I have reached to this point and gotten stuck. Can someone help me with how to move forward in this proof?

Answer 1

In metric space, it is separable if and only if it is second countable (having the countable neighborhood base).

Being second countable is hereditary, as intersection with the subspace.