Particular Point Space is First-Countable

Theorem
Let $T = \struct {S, \tau_p}$ be a particular point space.

Then $T$ is first-countable.

Proof
Let $x \in S: x \ne p$.

Consider the set $U_x = \set {x, p} \subseteq S$.

Now let $V \in \tau_p$ be an open set in $S$ such that $x \in V$.

So $x \in V$, by definition of $V$, and $p \in V$ as $V$ is open.

It follows directly that $U_x \subseteq V$.

So $\set {U_x}$ is a local basis at $x$ which is (trivially) countable.

If $x = p$ then $U_x = \set p$ and the same argument still applies.

So every element of $T$ has a countable local basis.

Hence $T$ is first-countable by definition.