Regular open algebra in ZF

Put $r(E)=\mathrm{int}\bar{E}$. This is regular open for any $E\subseteq X$: $r(E)$ is open and contained in the closed set $\bar{E}$, so $r(E)\subseteq r(r(E))\subseteq \mathrm{int}\bar{E}=r(E)$. Call $N$ nowhere dense if $\mathrm{int}\bar{N}=\mathrm{\varnothing }$. These sets form an ideal $\mathcal{N}$ of subsets: the empty set qualifies; subsets qualify by monotonicity; and if $N,M$ qualify, any nonempty open $W$ has a nonempty open part $W\setminus \bar{N}$, which in turn has a nonempty open part outside $\bar{M}$. Thus no nonempty open set lies in $\bar{N}\cup \bar{M}=\overline{N\cup M}$, proving the finite-union condition. A nowhere-dense set contains no nonempty open set.

For $E\subseteq X$ put $\mathrm{\partial }E=\bar{E}\setminus \mathrm{int}E$, and let $\mathcal{A}=\{E\subseteq X:\mathrm{\partial }E\in \mathcal{N}\}$. The identities $\mathrm{\partial }(X\setminus E)=\mathrm{\partial }E$ and $\mathrm{\partial }(E\cup F)\subseteq \mathrm{\partial }E\cup \mathrm{\partial }F$ follow respectively from $\overline{X\setminus E}=X\setminus \mathrm{int}E$ and the closure/interior union inclusions. They and $\mathrm{\partial }\mathrm{\varnothing }=\mathrm{\varnothing }$ show that $\mathcal{A}$ is closed under complements and finite unions, hence all finite set Boolean operations. It is therefore a Boolean algebra with those operations. If $N\in \mathcal{N}$, then $\mathrm{int}N=\mathrm{\varnothing }$ and $\mathrm{\partial }N=\bar{N}$ is nowhere dense, so $\mathcal{N}\subseteq \mathcal{A}$ and is an ideal of it. Every open $U$ belongs to $\mathcal{A}$: its closed boundary $\bar{U}\setminus U$ contains no nonempty open $W$, since $W\subseteq \bar{U}$ would force $W\cap U\mathrm{\ne }\mathrm{\varnothing }$.

For $E\in \mathcal{A}$, $\mathrm{int}E\subseteq r(E)\subseteq \bar{E}$ gives $E\mathrm{△}r(E)\subseteq \mathrm{\partial }E\in \mathcal{N}$. Also $r(E)\in \mathcal{A}$ by openness. If regular opens $U,V$ differ by a nowhere-dense set, the open set $U\setminus \bar{V}$ is a subset of that difference, so it is empty by step 1.1. Hence $U\subseteq \bar{V}$, and openness gives $U\subseteq \mathrm{int}\bar{V}=V$. Interchanging $U,V$ gives equality. More generally if both $U$ and $V$ differ from $E$ by nowhere-dense sets, then $U\mathrm{△}V\subseteq (U\mathrm{△}E)\cup (E\mathrm{△}V)$ is nowhere dense, so this uniqueness applies. Thus every class of $\mathcal{A}\mathrm{/}\mathcal{N}$ has exactly one regular-open representative, namely $r(E)$.

F2 and step 3.1 transport a Boolean algebra structure to $\mathrm{RO}(X)$. To identify its operations, first $U\cap V$ is regular: monotonicity gives $r(U\cap V)\subseteq r(U)\cap r(V)=U\cap V$, and the reverse inclusion follows from openness. Thus the transported meet is $U\cap V$. The transported join is $r(U\cup V)$, the representative of its set-theoretic union class. For the complement put $W=\mathrm{int}(X\setminus U)=X\setminus \bar{U}$. Since $W\subseteq X\setminus U$ and that latter set is closed, $r(W)\subseteq \mathrm{int}(X\setminus U)=W$; openness gives equality. Also $(X\setminus U)\mathrm{△}W=\bar{U}\setminus U$ is nowhere dense by step 2.1, so $W$ is the required complement representative. Bounds transport to $\mathrm{\varnothing },X$, which are regular. The transported order satisfies $U\le V$ iff $U\cap V=U$ iff $U\subseteq V$. In particular all distributive and complement laws hold by F2, rather than by a claim that regular opens are closed under ordinary unions.

For any $\mathcal{U}\subseteq \mathrm{RO}(X)$, $J=r(\bigcup \mathcal{U})$ is regular by step 1.1 and contains each $U\in \mathcal{U}$, since those sets are open subsets of $\overline{\bigcup \mathcal{U}}$. If a regular open $V$ contains each $U$, monotonicity gives $J\subseteq r(V)=V$, proving it is the least upper bound. Put $M=\mathrm{int}\bigcap \mathcal{U}$. For every $U\in \mathcal{U}$, $r(M)\subseteq r(U)=U$. Since $r(M)$ is open, it follows that $r(M)\subseteq \mathrm{int}\bigcap \mathcal{U}=M$; the reverse holds because $M$ is open. Thus $M$ is regular. Any regular-open lower bound is an open subset of $\bigcap \mathcal{U}$, hence lies in $M$, so $M$ is the greatest lower bound. If $\mathcal{U}$ is empty, these formulas give $J=r(\mathrm{\varnothing })=\mathrm{\varnothing }$ and $M=\mathrm{int}X=X$, directly verifying both bounds. If $X=\mathrm{\varnothing }$ there is just the one regular open $\mathrm{\varnothing }=X$, and the same proof gives the trivial complete Boolean algebra. No countable union of nowhere-dense sets or choice principle has been used. QED.