Open Extension of Double Pointed Countable Complement Topology is T4 Space

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

Let $D = \struct {\set {0, 1}, \vartheta}$ be the indiscrete topology on two points.

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

Let $\paren {T \times D}^*_{\bar p}$ be the open extension topology on $S \times \set {0, 1} \cup \set p$ where $p \notin S \times \set {0, 1}$.

Then $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space, and no other separation axioms are fulfilled.

That is, $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space or $T_5$ space.

Proof
From Open Extension Topology is T4 we have that $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space.

From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T \times D$ is not a $T_0$ space or a $T_5$ space.

From Condition for Open Extension Space to be $T_0$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space.

From Condition for Open Extension Space to be $T_5$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_5$ space.

From Open Extension Topology is not $T_1$, we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space.

From Open Extension Topology is not $T_3$ we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_3$ space.

Finally, from $T_2$ Space is $T_1$ Space, as $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space it is not a $T_2$ space.