## Any subspace of a separable metric space is separable

Let $X$ be a separable, metric space, $A \subset X$, and let $Q \subset X$ be countable and dense.  For each $q \in Q, n \in \mathbb{N}$, choose $a_n^q \in A \cap B_{1/n}(q)$ if such intersection is nonempty.  Then, $A^* = \bigcup_{n,q} \{ a_n^q \}$ is a countable subset of $A$.

Let $a \in A, 0 < \epsilon < 1$. Then choose $q \in B_{\epsilon / 4}(a) \cap Q$.  Now, choose $N$ such that $\epsilon / 4 \le 1 / N \le \epsilon / 2$.  Then, $B_{1/N}(q) \cap A \neq \emptyset$, so there exists $a^* \in A^*$, such that $d(a^*, q) < 1/N.$

Now $d(a^*,a) \le d(a^*,q) + d(q,a) < \epsilon / 2 + \epsilon / 4 < \epsilon$.  So $a^* \in B_\epsilon(a) \cap A$.  Hence, the closure of $A^*$ in $A$ is equal to $A$.