Open-dense and closed-nowhere-dense Baire forms are equivalent in ZF

From the ball definition of closure, $D$ is dense precisely when every ball about every point meets $D$. This is equivalent to meeting every nonempty open set: a point of such an open set has a ball inside it; conversely each ball is itself nonempty and open. It follows that $D$ is dense exactly when $\mathrm{int}(X\setminus D)=\mathrm{\varnothing }$, since a nonempty open subset of the complement is exactly an open set disjoint from $D$.

If $F$ is closed, $\bar{F}=F$, so $F$ is nowhere dense exactly when $\mathrm{int}(F)=\mathrm{\varnothing }$, exactly when $X\setminus F$ is dense. Its complement is open by closedness. Conversely, if $U$ is open dense, $F=X\setminus U$ is closed and has empty interior by the preceding test, hence is nowhere dense.

Assume (2). If a nonempty open $V$ were meagre, fix its one witness $V\subseteq {\bigcup }_n{N}_n$. Put $F_n=\overline{N_n}$ by the uniquely specified closure operation. Each $F_n$ is closed and has empty interior by nowhere density of $N_n$; by closedness its own closure equals itself. Thus the $F_n$ are closed nowhere dense. But $V\subseteq {\bigcup }_n{F}_n$ makes that union's interior nonempty, contradicting (2). This proves (3). The family of closures is defined from the given witness, without choosing decompositions.

Assume (3), and let $(F_n)$ be closed nowhere dense. If its union had nonempty interior $V$, this open set would be meagre, witnessed by the very sequence $(F_n)$, contrary to (3). Thus (2) follows.

Assume (3), and let $C$ be comeagre. If $C$ were not dense, the open-set test would give a nonempty open $V\subseteq X\setminus C$. A meagre witness for $X\setminus C$ also covers $V$, contradicting (3). Thus (4) follows. Conversely assume (4). If a nonempty open $V$ were meagre, $X\setminus V$ would be comeagre and hence dense, yet disjoint from $V$, a contradiction. Thus (4) implies (3).

Apply these complement correspondences term by term. For each sequence of closed nowhere dense $F_n$, the $U_n=X\setminus F_n$ are open dense and $X\setminus {\bigcup }_n{F}_n={\bigcap }_n{U}_n$. The union has empty interior exactly when this intersection is dense. Conversely, starting with any sequence of open dense $U_n$ and taking its closed nowhere dense complements gives the same identity. Thus (1) and (2) imply each other; the De Morgan index set is $\omega \mathrm{\ne }\mathrm{\varnothing }$.

These implications prove all four equivalences. They also cover $X=\mathrm{\varnothing }$: every set and every union or intersection under consideration is empty, hence dense with empty interior, and there is no nonempty open set. No choice axiom or completeness hypothesis was used.