Irreducible Hausdorff Space is Singleton

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

Then $T$ contains only one point.

Proof
Suppose $T$ has at least two points $x, y \in S$.

Because $T$ is irreducible, there are no two disjoint open sets such that $x$ is in one and $y$ is in the other.

This contradicts the fact that $T$ is Hausdorff.

Thus $T$ has only one point.