Double Pointed Finite Complement Topology is Compact

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

Let $T \times D$ be the double pointed topology on $T$.

Then $T \times D$ is compact.

Proof
The Finite Complement Topology is Compact.

Also, a Finite Topological Space is Compact.

So $D$, defined as the indiscrete space on a doubleton, is also compact.

The result follows directly from Tychonoff's Theorem.