Soberification of a topological space

The most important thing to note in order to prove that $\psi$ is continuous is the fact that opens of $pt(\Omega(X))$ are defined by the opens of $X$, i.e. the opens of $pt(\Omega(X))$ are precisely sets of the form $\phi(U) = \{V \in pt(\Omega(X)) \mid U$ is not contained in $V\}$, where $U \in \Omega(X)$ (see the wikipedia page on Stone duality). Take such an open $\phi(U)$ - we are going to consider $\psi^{-1}(\phi(U))$.

It holds that $x \in \psi^{-1}(\phi(U))$ if and only if $\overline{\{x\}}^c \in \phi(U)$, so if and only if $U$ is not contained in $\overline{\{x\}}^c$. It follows that $U \subset \psi^{-1}(\phi(U))$, since $x \notin \overline{\{x\}}^c$ for all $x \in U$. Suppose that $x \notin U$. Then $x \in U^c$, so $\overline{\{x\}} \subset U^c$ and $U \subset \overline{\{x\}}^c$, so $x \notin \psi^{-1}(\phi(U))$. Thus $U = \psi^{-1}(\phi(U))$, which shows that $\psi$ is continuous.

The function $\psi$ is onto if and only if every closed irreducible subset of $X$ is the closure of a point $x \in X$. This holds if $X$ is an affine scheme, for example. I was unable to further specify when $\psi$ is onto.