Asaf's argument : (ZF) If $\mathbb{R}^k$ is a countable union of closed sets, then at least one has a nonempty interior
Let $X$ be a separable complete metric space. Let $D$ be a countable debse subset. Let $F$ be a closed subset with empty interior.
Here, how do i show that $D\setminus F$ is dense?
Ive been trying to figure it out for a day, and still stuck here..