Hausdorff implies sober

Statement
Any Hausdorff space is a sober space.

Hausdorff space
A topological space $$X$$ is said to be Hausdorff if, for any two points $$x,y$$ in $$X$$, there exist open subsets $$U,V$$ of $$X$$ such that $$x \in U, y \in V$$ and $$U \cap V$$ is empty.

Sober space
A topological space $$X$$ is termed sober if the only irreducible closed subsets of $$X$$ are the closures of one-point subsets.

Proof
We execute this proof by showing that, for a Hausdorff space, the only irreducible closed subsets are the one-point subsets. Note that this in particular shows that the only irreducible closed subsets are the closures of one-point subsets.

Given: A topological space $$X$$, an irreducible closed subset $$A$$ of $$X$$.

To prove: $$A$$ is a one-point subset.

Proof: Since $$A$$ is irreducible, it is nonempty, so it has at least one point. We show that if $$A$$ has two points, we obtain a contradiction.

Suppose $$x,y \in A$$ are two distinct points. By the definition of Hausdorffness, there exist open subsets $$U,V$$ of $$X$$ such that $$U \ni x, V \ni y$$, and $$U \cap V = \emptyset$$. Then, the sets $$A \setminus U$$ and $$A \setminus V$$ are both closed subsets, their union is $$A$$, and they are both proper subsets of $$A$$ since $$x \notin A \setminus U$$ and $$y \notin A \setminus V$$. Thus, we have written $$A$$ as a union of two proper closed subsets, contradicting the assumption of irreducibility.