Countable Discrete Space is not Weakly Countably Compact

Theorem
Let $T = \struct {S, \tau}$ be a countable discrete space.

Then $T$ is not weakly countably compact.

Proof
Let $A \subseteq S$ be an infinite subset of $S$.

Then as Set in Discrete Topology is Clopen, $A$ is closed in $T$.

From Closed Set Equals its Closure, $A = A^-$ where $A^-$ is the closure of $A$.

From the definition, $x$ is a limit point of $A$ if it belongs to the closure of $A$ but is not an isolated point of $A$.

But then we have Topological Space is Discrete iff All Points are Isolated.

So $A$ has no limit points in $T$.

So $T$ is not weakly countably compact.