Fortissimo Space is not Weakly Countably Compact

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

Then $T$ is not weakly countably compact.

Proof
It suffices to show that $T$ has an infinite subset without limit points.

Consider the set $S \setminus \set p$.

For any $x \in S$:
 * $\paren {S \setminus \paren {S \setminus \set p}} \cup \set x = \set p \cup \set x = \set {p, x}$

By definition, $x$ is a limit point of $S \setminus \set p$ $\set {p, x}$ is not a neighborhood of $x$.

By, $\set {p, x}$ is open in $T$.

Hence it is a open neighborhood of $x$.

Therefore $x$ is not a limit point of $S \setminus \set p$.

Since $x$ is arbitrary, $S \setminus \set p$ has no limit points.

Hence $T$ is not weakly countably compact.