T1/2 Space is T0 Space

Theorem
Let $T = \left({S, \tau}\right)$ be a $T_{\frac 1 2}$ topological space.

Then $T$ is $T_0$ space.

Proof
By Characterization of T0 Space by Closures of Singletons it suffices to prove that
 * $\forall x, y \in S: x \ne y \implies x \notin \left\{{y}\right\}^- \lor y \notin \left\{{x}\right\}^-$

where $\left\{{y}\right\}^-$ denotes the closure of $\left\{{y}\right\}$

Let $x, y$ be points of $T$ such that
 * $x \ne y$

Aiming for a contradiction suppose
 * $x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$

We will prove that
 * $x \notin \left\{{x}\right\}'$

where $ \left\{{x}\right\}'$ denotes the derivative of $\left\{{x}\right\}$

Aiming for a contradiction suppose
 * $x \in \left\{{x}\right\}'$

As $S$ is open by definition of topological space by Characterization of Derivative by Open Sets:
 * $\exists z \in S: z \in \left\{{x}\right\} \cap S \land z \ne x$

Bu definition of intersection:
 * $z \in \left\{{x}\right\}$

This contradicts $z \ne x$ by definition of singleton

Thus $x \notin \left\{{x}\right\}'$

We will prove that:
 * $(1): \quad \lnot \forall G \in \tau: y \in G \implies \left\{{x}\right\} \cap G \ne \varnothing$

Aiming for a contradiction suppose
 * $\forall G \in \tau: y \in G \implies \left\{{x}\right\} \cap G \ne \varnothing$

As sublemma we will show that
 * $\forall U \in \tau: y \in U \implies \exists r \in S: r \in \left\{{x}\right\} \cap U \land y \ne r$

Let $U \in \tau$ such that
 * $y \in U$

Then by assumption
 * $\left\{{x}\right\} \cap U \ne \varnothing$

By definition of empty set:
 * $\exists z: z \in \left\{{x}\right\} \cap U$

By definition of intersection:
 * $z \in \left\{{x}\right\}$

Then by definition of singleton:
 * $z = x \ne y$

Thus:
 * $\exists r \in S: r \in \left\{{x}\right\} \cap U \land y \ne r$

Then by Characterization of Derivative by Open Sets:
 * $y \in \left\{{x}\right\}'$

By definition of relative complement:
 * $y \notin \complement_S \left({\left\{{x}\right\}'}\right) \land x \in \complement_S \left({\left\{{x}\right\}'}\right)$

By definition of $T_{\frac 1 2}$ space:
 * $\left\{{x}\right\}'$ is closed

By definition of closed set:
 * $\complement_S \left({\left\{{x}\right\}'}\right)$ is open

By $x \in \left\{{y}\right\}^-$ and by Condition for Point being in Closure:
 * $\left\{{y}\right\} \cap \complement_S \left({\left\{{x}\right\}'}\right) \ne \varnothing$

Then by definition of empty set:
 * $\exists z: z \in \left\{{y}\right\} \cap \complement_S \left({\left\{{x}\right\}'}\right)$

By definition of intersection:
 * $z \in \left\{{y}\right\} \land z \in \complement_S \left({\left\{{x}\right\}'}\right)$

By definition of singleton:
 * $z = y$

Thus $z \in \complement_S \left({\left\{{x}\right\}'}\right)$ contradicts $y \notin \complement_S \left({\left\{{x}\right\}'}\right)$

This ends the proof of $(1)$.

Then by $(1)$ ans Condition for Point being in Closure
 * $y \notin \left\{{x}\right\}^-$

This contradicts $y \in \left\{{x}\right\}^-$