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.