Either-Or Topology is not T3

Theorem
Let $T = \struct {S, \tau}$ be the either-or space.

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

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

Let $x = \dfrac 1 2$.

Let $U \in \tau$ such that $\set 0 \subseteq U$.

Then by definition $\openint {-1} 1 \subseteq U$ 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.