Fortissimo Space is T5

Theorem
Let $T = \left({S, \tau_p}\right)$ be a Fortissimo space on an infinite set $S$.

Then $T$ is a $T_5$ space.

Proof
Let $A, B \in \tau_p$ such that $A$ and $B$ are separated.

If $p \notin A$ and $p \notin B$ then $A$ and $B$ are both open.

Otherwise $p$ must be in exactly one of them, because if $p$ were in both they could not be separated.

Suppose WLOG that $p \in A$.

Then $p \notin B$ so $B$ is open.

Now suppose $B$ were not closed.

Then $B$ is uncountable by definition of Fortissimo space.

But every open set not containing $p$ is co-countable in $S$.

So $p$ would be in $B^-$, where $B^-$ is the closure of $B$.

But then $p \in A$, by hypothesis.

So $A \cap B^- \ne \varnothing$ and so $A$ and $B$ are not separated.

So $B$ must be closed.

Therefore $\complement_S \left({B}\right)$ is an open set such that $A \subseteq \complement_S \left({B}\right)$, and the $T_5$ separation axiom is satisfied.