Particular Point Topology with Three Points is not T4

Theorem
Let $T = \left({S, \tau_p}\right)$ be a particular point space such that $S$ is not a singleton or a doubleton.

That is, such that $S$ has more than two distinct elements.

Then $T$ is not a $T_4$ space.

Proof
We have that there are at least three elements of $S$.

So, consider $x, y, p \in S: x \ne y, x \ne p, y \ne p$.

Then $X = \left\{{x}\right\}, Y = \left\{{y}\right\}$ are closed in $T$ and $X \cap Y = \varnothing$.

Suppose $U, V \in \tau_p$ are open sets in $T$ such that $X \subseteq U, Y \subseteq V$.

But as $p \in U, p \in V$ we have that $U \cap V \ne \varnothing$.

So $T$ is not a $T_4$ space.

Also see

 * Sierpiński Space is $T_4$
 * Sierpiński Space is $T_5$

Mistakes in Sources
See : $\text{II}: \ 8 - 10: \ 4$ where it is stated that:
 * Non-Trivial Particular Point Topology is not T4