Countable Complement Space is not Weakly Countably Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a countable complement topology on an uncountable set $S$.

Then $T$ is not weakly countably compact.

Proof
By definition, $T$ is weakly countably compact every infinite subset of $S$ has a limit point in $S$.

Let $H \subseteq S$ be a countable set.

From Limit Points of Countable Complement Space, it contains all its limit points.

Let $x \in H$.

$\relcomp S {\relcomp S H \cup \set x} = H \setminus \set x$ is countable.

Then $\relcomp S H \cup \set x$ is an open neighborhood of $x$.

However $H \cap \paren {\paren {\relcomp S H \cup \set x} \setminus \set x} = H \cap \relcomp S H = \O$.

So $x$ is not a limit point of $H$.

Since $x$ is arbitrary, there are no limit points in $H$.

This shows that $T$ is not weakly countably compact.