Either-Or Topology is not T3

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

Then $T$ is not a $T_3$ space.

Proof
Consider the set $\left\{{0}\right\}$ which is closed from Closed Sets of Either-Or Topology.

Let $x = \dfrac 1 2$.

Let $U \in \tau$ such that $\left\{{0}\right\} \subseteq U$.

Then by definition $\left({-1 \,.\,.\, 1}\right) \subseteq U$ and so $x \in U$.

So we have found a closed set of $T$ and a point in $S$ which do not satisfy the conditions for $T$ to be a $T_3$ space.

Hence the result.