# Any subspace of a separable metric space is separable

Proposition. Let $X$ be a separable metric space, $A$ be a subpace of $X$, then $A$ is also separable.,

Proof. let $Q\subset X$ be a countable and dense subset.  For each $q\in Q$ and $n\in\mathbf{N}$, choose $a_{q,n}\in A\cap B(q,1/n)$ if such intersection is nonempty.  Then, $A^*=\bigcup_{q,n} \{a_{q,n}\}$  is a countable subset of  $A$.

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

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