Double Pointed Finite Complement Topology fulfils no Separation Axioms

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 not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.

Proof
From Double Pointed Topology is not $T_0$, we have that $T \times D$ is not a $T_0$ space.

From $T_1$ Space is $T_0$ Space, $T \times D$ is not a $T_1$ space.

From $T_2$ Space is $T_1$ Space, $T \times D$ is not a $T_2$ space.

As the Finite Complement Space is Irreducible, $T$ has no disjoint open sets.

So for any closed set $F$ and $x \in S: x \notin F$ there is no pair of disjoint open sets of $T$ one of which contains $F$ and the other containing $x$.

So by definition $T$ is not a $T_3$ space.

For exactly the same reason $T$ is not a $T_4$ space.

Finally, from $T_5$ Space is $T_4$ Space, $T \times D$ is not a $T_5$ space.