Excluded Point Space is T0

Theorem
Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Then $T$ is a $T_0$ (Kolmogorov) space.

Proof 1
If $T$ is a trivial space, i.e. such that $S = \left\{{p}\right\}$, the result holds vacuously - there are no two distinct points in $T$.

Now suppose $T$ is not trivial.

Then $\exists x \in S: x \ne p$.

Now we have that $\left\{{x}\right\} \subseteq T$ is open in $T$ such that $p \notin \left\{{x}\right\}$ but $x \in \left\{{x}\right\}$.

Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.

Then we have that (for example) $\left\{{x}\right\} \subseteq T$ is open in $T$ such that $x \in \left\{{x}\right\}$ but $y \notin \left\{{x}\right\}$

Hence the result.

Proof 2
We have:
 * Excluded Point Topology is Open Extension Topology of Discrete Topology


 * Discrete Space Satisfies All Separation Properties (including being a $T_0$ space)

Then by Condition for Open Extension Space to be $T_0$ Space, as a discrete space is $T_0$ then so is its open extension.