Excluded Point Space is T5

Theorem
Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Then $T$ is a $T_5$ space.

Proof
Let $A, B \subseteq S$ such that:
 * $A^- \cap B = A \cap B^- = \varnothing$

where $A^-$ denotes the closure of $A$.

Every closed set of $T$ contains $\left\{{p}\right\}$.

From Limit Points in Excluded Point Space and the definition of closure:
 * $A^- = A \cup \left\{{p}\right\}, B^- = B \cup \left\{{p}\right\}$

So if $A^- \cap B = A \cap B^- = \varnothing$ then $p \in A^-$ and $p \in B^-$.

Thus, $p \notin A$ and $p \notin B$ and so by definition $A$ and $B$ are open sets in $T$.

It follows that $A \cap B = \varnothing$.

Hence the result, as we have shown that:
 * $\forall A, B \subseteq X, A^- \cap B = A \cap B^- = \varnothing: \exists U, V \in \vartheta: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

where $U = A$ and $V = B$.