Is a separable and metrizable space second countable?

Question

I believe the answer to this question is yes. I was hoping someone would critique my logic.

Let X be a separable and metrizable space. Then it has a countable dense subset A. Let B be a basis for X.

For every a in A, choose a basis element U in B such that a is in U (if such exists).

This forms a countable collection V. Suppose, for a contradiction, there is an element x that is in no element of V. Then x is in a basis element U. U must contain an element of A since A is dense. But this U is not in our collection V, a contradiction.

Answer 1

Separability and second countability coincide for metrizable spaces.

For a proof see e.g.

here A metric space is separable iff it is second countable