Irreducible and Hausdorff implies one-point space

Statement
Suppose $$X$$ is a non-empty topological space that is both an uses property satisfaction of::irreducible space and a uses property satisfaction of::Hausdorff space. Then, $$X$$ must be a one-point space.

Similar facts

 * Ultraconnected and T1 implies one-point space

Proof
We prove a somewhat modified form: we prove that an Hausdorff space containing two distinct points cannot be irreducible.

Given: A non-empty topological space $$X$$ that is Hausdorff and has at least two points. Say $$x,y$$ are two distinct points of $$X$$.

To prove: $$X$$ is not irreducible, i.e., it contains two non-empty open subsets with empty intersection.

Proof: Use the definition of Hausdorff to find non-empty open subsets $$U \ni x, V \ni y$$ that have empty intersection. This completes the proof.