Let $\Omega$ be a topological space. If $A\subseteq\Omega$, then $$\overline A=\bigcap_{\substack{B\supseteq A\\B\text{ is closed}}}B\tag1.$$ If the topology on $\Omega$ is generated by a metric $d$, then $$B_\varepsilon:=\left\{x\in\Omega:d(x,A)<\varepsilon\right\}$$ is open for all $\varepsilon>0$ and $$\overline A=\bigcap_{\varepsilon>0}B_\varepsilon=\bigcap_{n\in\mathbb N}B_{\frac1n}\tag2.$$
By $(2)$, the closure of an arbitrary subset of a metric space can be written as the (countable) intersection of open neighborhoods. Does the same hold in an arbitrary topological space?
Best Answer
No, the property that every closed set is a countable intersection of open sets (which is what your property comes down to) is called being perfectly normal (usually in combination with being $T_4$ as well). I've also seen it called a $G_\delta$ space. All metric spaces have this property, but not all spaces do:
The Michael line, and the lexicographically ordered square are some examples of non-metrisable but nice (herditarily normal etc) spaces that do not satisfy this. And $[0,1]^I$ with uncountable $I$ is compact Hausdorff where even singleton sets are not $G_\delta$'s.