Countable Excluded Point Space is Second-Countable/Proof 1

Proof
Consider the set $\BB$ defined as:
 * $\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$

From Basis for Excluded Point Space, $\BB$ 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.