Either-Or Topology is T0

Theorem
Let $T = \left({S, \tau}\right)$ be the either-or space.

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

Proof
Let $x, y \in S$ such that $x \ne y$.

, let $x \ne 0$.

Then $U = \left\{{x}\right\}$ is open in $T$ from the definition of the either-or topology.

So $\exists U \in \tau: x \in H, y \notin H$.