Particular Point Space is First-Countable

Theorem
Let $T = \left({S, \tau_p}\right)$ be a particular point space.

Then $T$ is first-countable.

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

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

Now let $V \in \vartheta_p$ 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 $\left\{{U_x}\right\}$ is a local basis at $x$ which is (trivially) countable.

If $x = p$ then $U_x = \left\{{p}\right\}$ and the same argument still applies.

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

Hence $T$ is first-countable by definition.