Suppose $\mathcal{A}$ is a non-empty upward-closed family (i.e. $A\in \mathcal{A}$ and $A\subseteq B$ implies $B\in\mathcal{A}$) of non-empty subsets on a non-empty set $X$.
In ZFC, does there exist a topology $\tau$ on $X$ such that dense subsets $\mathfrak{D}(\tau)$ of $(X, \tau)$ are precisely elements of $\mathcal{A}$, that is $\mathfrak{D}(\tau) = \mathcal{A}$?
Since all $A\in\mathcal{A}$ are dense in $(X, \{\emptyset, X\})$ and if $\tau_i$ is chain of topologies such that all $A\in \mathcal{A}$ are dense in $(X, \tau_i)$, and if $\tau = \{\bigcup_i V_i : V_i\in\tau_i\}$ is the topology generated by $\bigcup_i\tau_i$, then all $A\in\mathcal{A}$ are dense in $(X, \tau)$. From Zorn's lemma there exists a maximal topology such that all $A\in\mathcal{A}$ are dense.
In fact in the above argument $\mathfrak{D}(\tau) = \bigcap_i \mathfrak{D}(\tau_i)$.
Taking a maximal element doesn't seem to be enough to prove this though. I was trying to modify it but all my attempts failed.
This is not the case without further assumptions of some sort.
As a counterexample, suppose $X$ has at least three elements, and let $\mathcal A$ be the family of subsets with at least two elements. Then on the one hand, every singleton must be closed (i.e., $X$ is $T_1$), since otherwise we have $\overline{\{x\}}\in \mathcal A$, and $\overline{\{x\}}\neq X$, but also $\overline{\overline{\{x\}}}=X$, contradicting the fact that in any topological space, $\overline{\overline{S}}=\overline{S}$ for all $S\subseteq X$.
On the other hand, we then have that every finite set is closed, so two element sets cannot be dense.