Regular opens form base for compact Hausdorff space I suspect that the regular open sets, i.e. the sets such that $U = \operatorname{Int}(\operatorname{Cl}(U))$, of a compact Hausdroff space form a base for the topology. But I cannot find it in literature. Does anybody know if this is even true?
 A: Yes, because compact Hausdorff spaces are regular. 

Regular spaces have a base of regular open sets.

Suppose $x$ is a point and $U$ is a neighborhood of $x$. We just need to find a regular open subset $W$ such that $x\in W \subseteq U$. By applying regularity to $x$ and $U^c$, we obtain an open set $V$ such that $
\newcommand{\Int}{\operatorname{Int}}\newcommand{\Cl}{\operatorname{Cl}}
x\in V \subseteq \Cl(V) \subseteq U$.
Define $W:=\Int(\Cl(V))$. Then $W$ is open and $x\in V \subseteq W\subseteq \Cl(V) \subseteq U$, so it only remains to show that $W$ is regular open.
To do this, note that
$$
\Int(\Cl(W))=\Int(\Cl(\Int(\Cl(V))))=\Int(\Cl(V))=W,
$$
where the second equality above uses the following fact.

If $A$ is a subset of a topological space, then $\Int(\Cl(\Int(\Cl(A))))=\Int(\Cl(A))$.

A search of the site led me to this by Henno Brandsma. See fact (p):
We have $\Int(\Cl(A))\subseteq \Cl(\Int(\Cl(A)))$, so by taking interiors we obtain
$$
\Int(\Cl(A))=\Int(\Int(\Cl(A))) \subseteq \Int(\Cl(\Int(\Cl(A)))).
$$
For the reverse inclusion, we have $\Int(\Cl(A))\subseteq\Cl(A)$, so taking closures yields
$$
\Cl(\Int(\Cl(A)))\subseteq \Cl(\Cl(A))=\Cl(A).
$$
Now applying interiors gives
$$
\Int(\Cl(\Int(\Cl(A)))) \subseteq \Int(\Cl(A)),
$$
establishing the equality.
