Any sober space is $T_0$

Question

Recall that a sober space is a topological space $X$ with the property that each closed irreducible set is the form $cl \{x\} $ for some $x \in X$.

How to prove that each sober space is necessarilly $T_0$?

Answer 1

Your characterization of sober spaces is slightly different than the usual characterization, which is that every closed irreducible set is of the form $\overline{\{x\}}$ for some unique $x\in X$, where we use an overline to denote the topological closure. (See eg here or here, where this is the definition given.)

With this definition the result is clear. Indeed, suppose $X$ is not $T_0$; then there are $x,y\in X$ such that every open set of $X$ either contains both $x$ and $y$ or contains neither of them. Taking complements, this means that every closed set of $X$ contains either both or neither of $x$ and $y$, and so $\overline{\{x\}}=\overline{\{y\}}$ is a closed irreducible set expressable as the closure of two distinct points of $X$, whence $X$ is not sober.

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If we get rid of the uniqueness demand in our definition, then there are sober spaces that are not $T_0$. For instance, consider the space $\{\star,\bullet\}$ with the indiscrete topology. This is in some sense a minimal example, as it embeds into every non-$T_0$ space.