# In which spaces,$F$ is irreducible iff $F=\overline{\{x\}}$, for all closed $F$?

I'm wondering which spaces $X$ meet the following condition:

For any closed $F\subseteq X$, $F$ is irreducible if and only if $F=\overline{\{x\}}$ for some $x\in X$.

Where $F$ is irreducible if and only if for all closed sets $F_1,F_2$ such that $F\subseteq F_1\cup F_2$, we have that $F\subseteq F_1$ or $F\subseteq F_2$.

Thanks.

A point $x\in F$ whose closure is $F$ is called a generic point of $F$ (note that the closure of an irreducible subspace of topological space is always irreducible). A sober topological space is one in which every irreducible closed subset has a unique generic point. Any such space has the property you ask about (the existence of a generic point for each irreducible closed subset). Schemes (e.g. spectra of rings) are sober.
The uniqueness in the definition of sobriety is equivalent to requiring that for distinct points $x\neq y$ in the space, there is either a neighborhood of $x$ not containing $y$, or a neighborhood of $y$ not containing $x$ (I believe this is called $T_0$ or Kolmogorov); more succinctly, $\overline{\{x\}}\neq\overline{\{y\}}$. The indiscrete topology on a set with two elements $X=\{x,y\}$ is such that every irreducible closed set has a generic point, but the generic point is not unique (the space is not $T_0$). So this is another example of a space having the property you ask about, but which is not sober.