I'm not talking about Stone spaces! A Stonean space is a compact Hausdorff extremally disconnected space i.e. a compact Hausdorff space such that for all open $U$, its closure $\overline{U}$ is also open.
I've been thinking about applications of regular open sets in topology, and I came up with the following:
If $X$ is a topological space then the set $\text{RO}(X)$ of regular open sets of $X$ forms a complete Boolean algebra. This Boolean algebra induces a Stonean space $S(X)$ by Stone duality. Thus to any topological space we can assign a Stonean space $S(X)$.
If $X$ is a Stonean space, then $\text{RO}(X)$ is a Boolean algebra of clopen sets of $X$, so that Stone duality tells us $S(X)$ gives us $X$ back.
If $f:X\to Y$ is a continuous map, then $f^{-1}(\text{RO}(Y))\subseteq \text{RO}(X)$ is not true in general (see comments below). Thus I want to restrict to continuous open maps. In this case, if $U\in\text{RO}(Y)$ then $$\text{int }\overline{f^{-1}(U)} = \text{int }f^{-1}(\overline{U}) = f^{-1}(\text{int }\overline{U})$$ see also this post by Paul Frost. So $f^{-1}$ induces a map from $\text{RO}(Y)$ to $\text{RO}(X)$. Then the Stone duality induces a map $S(f):S(X)\to S(Y)$.
Now, using this, does that mean that Stonean spaces are a reflective subcategory of topological spaces with continuous open maps?
Best Answer
No, Stonean spaces are not reflective in all topological spaces with continuous open maps. Consider an arbitrary Stone space $X$. Note that every continuous map $X\to\{0,1\}$ is automatically open. So, if a reflection $i:X\to Y$ of $X$ into Stonean spaces existed, every continuous map $X\to\{0,1\}$ would factor through $i$ and so $i$ would be injective since these maps separate points of $X$. But since $i$ is open and $X$ is compact, this would mean that $i$ is an embedding of $X$ as a clopen subspace of $Y$, so $X$ itself must be Stonean. So any non-Stonean Stone space does not have a Stonean reflection.
(The problem with the construction you propose is that there is not actually a canonical continuous open map $X\to S(X)$. Very concretely, a point of $S(X)$ is an ultrafilter on $RO(X)$. But a point of $X$ doesn't naturally give an ultrafilter on $RO(X)$, since if $U\in RO(X)$, then $U$ and its complement in $RO(X)$ may not actually cover all of $X$, so the filter on $RO(X)$ given by which regular open sets contain a given point of $X$ may not be an ultrafilter.)