Second-Countable Space is First-Countable

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

Then $T$ is also both first-countable and separable.

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


 * If we consider the entire set $X$ as an open set, then we see that $X$ is a neighborhood of all points.

As $T$ has a countable basis, then (trivially) every point in $T$ has a countable local basis.

So a second-countable space is trivially first-countable.


 * 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 $x \in S$ such that $x \notin H$.

As $\mathcal B$ is a basis for $\vartheta$, it follows that $U$ is the union of elements of $\mathcal B$.

Then $\forall U \in \vartheta: x \in U$ 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 $H$ that 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.