Second-Countable Space is Separable

Theorem
Let $T = \left({S, \tau}\right)$ be a second-countable topological space.

Then $T$ is also a separable space.

Proof
By definition, there exists a countable basis $\mathcal B$ for $\tau$.

Using the axiom of countable choice, we can obtain a choice function $\phi$ for $\mathcal B \setminus \left\{{\varnothing}\right\}$.

Define:
 * $H = \left\{{\phi \left({B}\right): B \in \mathcal B \setminus \left\{{\varnothing}\right\}}\right\}$

By Image of Countable Set under Mapping is Countable, it follows that $H$ is countable.

It suffices to show that $H$ is everywhere dense in $T$.

Let $x \in U \in \tau$.

By Equivalence of Definitions of Analytic Basis, there exists a $B \in \mathcal B$ such that $x \in B \subseteq U$.

Then $\phi \left({B}\right) \in U$, and so $H \cap U$ is non-empty.

Hence, $x$ is an adherent point of $H$.

By Equivalence of Definitions of Adherent Point, it follows that $x \in H^-$, where $H^-$ denotes the closure of $H$.

Therefore, $H^- = S$, and so $H$ is everywhere dense in $T$ by definition.