Irreducible Hausdorff Space is Singleton

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is irreducible.

Then $T$ is not a $T_2$ (Hausdorff) space.

Proof
Let $x, y \in S$.

Let $T = \left({S, \tau}\right)$ be irreducible.

Then:
 * $\forall U_1, U_2 \in \tau: U_1, U_2 \ne \varnothing \implies U_1 \cap U_2 \ne \varnothing$

So trivially there are no two disjoint open sets such that $x$ is in one and $y$ is in the other.

Hence the result by definition of $T_2$ (Hausdorff) space.