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 infinite subsets on an infinite 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 $\mathfrak{D}(\tau) = \mathcal{A}$?
M W have shown that the above is false if elements of $\mathcal{A}$ can be finite, since given $n\geq 2$, the family $$\mathcal{A} = \{Y\subseteq X : |Y| \geq n\}$$ where $|X| \geq n+1$ doesn't correspond to a family of dense subsets, since that would imply $X$ is $T_1$ yet we can find a proper dense subset of $X$ with $n$ elements.
This is a follow up question, is it still false if all elements of $\mathcal{A}$ are infinite?
Yes, it is still false. Consider $X=\mathbb N\times \mathbb N$, and let $\mathcal A$ be all sets that contain infinitely many members of at least two rows. Then $S=\mathbb N \times\{0\}\notin \mathcal A$, so $\overline{S}\neq X$. Since $\overline{\overline{S}}=\overline{S}$, we also have $\overline{S}\notin \mathcal A$. Similarly $\overline{T}\notin \mathcal A$ for $T=\mathbb N \times \{1\}$. Therefore $\overline{S}$ and $\overline{T}$ each contain only finitely many members in all but one row. It follows that $\overline{S\cup T}=\overline{S}\cup\overline{T}\neq X$, but this contradicts that $S\cup T\in \mathcal A$.