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If an open set meets $\overline{A}$, it meets $A$. Thus if two open sets meet the closure of an irreducible set, they both meet $A$, their intersection meets $A$ by irreducibility, and therefore it also meets $\overline{A}$. This proves irreducibility of the closure. A singleton is irreducible, so its closure is an irreducible closed subset.

In a Hausdorff space, two distinct points have disjoint open neighborhoods. An irreducible set cannot contain both, so every nonempty irreducible closed subset is a singleton, which is closed and has a unique generating point. Hence every Hausdorff space is a sober topological space. An infinite set with the cofinite topology is itself irreducible, because any two nonempty open sets meet, but every singleton is closed. Its whole space is therefore not the closure of a singleton, proving that it is not sober.

For the proposed sobrification, let $\hat{X}$ consist of nonempty irreducible closed subsets. The basic operations areThe last identity uses irreducibility of each member of $\hat{X}$. Hence these sets already form a topology. The map $U\mapsto \hat{U}$ is injective: $\overline{\{x\}}$ meets an open $U$ exactly when $x\in U$, so equality of the hats forces equality of the original opens. It is surjective onto the topology just constructed by definition.

Now let $C$ be a nonempty irreducible closed subset of $\hat{X}$. Its complement is $\hat{W}$ for a unique open $W\subseteq X$. Put $F=X\setminus W$. ThenIf opens $U,V$ meet $F$, choose points in those intersections. Their singleton closures lie in $C$, so $\hat{U},\hat{V}$ meet $C$. Irreducibility of $C$ forces $\widehat{U\cap V}$ to meet $C$, and hence $U\cap V$ meets $F$. Thus $F$ is nonempty, closed and irreducible, so $F$ is itself a point of $\hat{X}$.

The closure of this point in $\hat{X}$ is exactly $\{A:A\subseteq F\}$: inclusion makes every open meeting $A$ meet $F$, whereas a point of $A\setminus F$ is separated by the open $X\setminus F$. Hence $C=\overline{\{F\}}$. Any other point $G$ with the same closure satisfies $G\subseteq F$ and $F\subseteq G$, forcing $F=G$. This proves the constructed space is sober.