Let $(X,d)$ be a metric space, and $E\subset X$ a generic closed set. Under what kind of condition is it true that there exists a sequence $\{x_n\}\subset X$ such that $\overline{\{x_n\}}=E$ (meaning the closure of $\{x_n\}$)?
I proved that $(X,d)$ must be separable, and I managed to prove that the statement is true for $(\mathbb{R},|\dot{}|)$. Is it true for every metric separable space? If not, what are the additional conditions that we need?
By definition $E$ is separable iff $E$ is the closure of some countable set $\{x_1,x_2,...\}$. Every (closed) subset of a metric space $X$ is separable iff $X$ itself is separable. So separability of $X$ is enough.
Let $\{x_n\}$ be a countable dense subset of $X$. Consider the open balls $B(x_n,\frac 1 m)$. If this intersects $E$ pick one point $y_{nm}$ in it. Otherwise ignore this ball. I leave it to you to verify that $\{y_{nm}\}$ is a countable dense subset of $E$.