Second-Countable Space is Separable

Theorem
Let $T = \left({X, \vartheta}\right)$ be a topological space which is second-countable.

Then $T$ is also a separable space.

Proof
$T$ is second-countable iff its topology has a countable basis.

Let $\mathcal B$ be a countable basis for $\vartheta$.

Consider the set $H \subseteq X$ defined by selecting one point from each element of $\mathcal B$.

Thus:
 * $\forall B \in \mathcal B: H \cap B \ne \varnothing$

Let $x \in X \setminus H$.

As $\mathcal B$ is a basis for $\vartheta$, it follows that any open neighborhood $U$ of $x$ is a union of elements of $\mathcal B$.

Thus for all such neighborhoods $U$ it follows that $\exists B \in \mathcal B: x \in B$.

But as $H \cap B \ne \varnothing$ it follows that at least one element of $H$ (which therefore is not $x$) is in $U$.

Hence by definition $x$ is a limit point of $H$.

Again by definition, $H$ is a countable everywhere dense subset of $T$.

So $T$ is by definition separable.