If I have a point $x \in S$ where $S$ is some Polish Space with metric $d$, a closed set $F \subset S$, and $inf\;\{d(x,f)|f\in F\}=0$, then why can we say that there exists $f' \in F$ such that $d(x,f')=0$?
If we had some kind of compactness it would be easy but my topology class is ages ago and I can´t see a solution right now.
For each $n > 0$, there is $y_n \in F$ with $d(y_n,x) \leq 1/n$. So by definition, $y_n \rightarrow x$, so that $x$ is in the closure of $F$. Thus ($F$ closed) $x \in F$.