Countable Excluded Point Space is Second-Countable/Proof 1

Proof
Consider the set $\mathcal B$ defined as:
 * $\mathcal B = \left\{{\left\{{x}\right\}: x \in S \setminus \left\{{p}\right\}}\right\} \cup \left\{{S}\right\}$

From Basis for Excluded Point Space, $\mathcal B$ is a basis for $T$, and trivially has the same cardinality as $S$.

So by definition, if $S$ is countable, then $T$ is second-countable.