The interior of an arbitrary intersection of regularly open sets is regularly open

Part of Exercise 4.16 (pp. 17-18) of W.A.J. Luxemburg & A.C. Zaanen's Riesz Spaces: Volume I (Google books link) is to prove

... the interior of an arbitrary intersection of regularly open sets is regularly open.

Can you suggest me how this holds?

Definition. For a topological space $X$ an open set $A$ is regularly open if interior of the closure of $A$ is again $A$, that is if $A= \operatorname{int} \overline{A}$.

Facts:

1. If we define $A^* = \operatorname{int} \overline{A}$, then for any open set $A$ we have $A\subseteq A^* \subseteq \overline{A}$.
2. The union of two regularly open sets may not be regularly open. Finite intersections of regularly open set are regularle open.

The arbitrary intersection of regularly open set is not necessarily regularly open, but note that here we are considering the interior of the intersection.

Let $\{A_\alpha\}$ be a family of regular open sets and $A=\operatorname{int}\bigcap A_\alpha$. Then $\operatorname{int} \overline{A} \subset \operatorname{int} \overline{A_\alpha}=A_\alpha$ for each $A_\alpha$, so $\operatorname{int} \overline{A} \subset \bigcap A_\alpha$. Since the set $\operatorname{int} \overline{A}$ is open, we see that $\operatorname{int} \overline{A} \subset \operatorname{int}\bigcap A_\alpha=A$.