 Solution

If $x\notin \overline{A}$, some closed superset $F$ of $A$ misses $x$. Its complement is an open set containing $x$ and disjoint from $A$. Conversely, if an open set $U$ contains $x$ and misses $A$, the closed set $X\setminus U$ contains $A$ and misses $x$, so $x\notin \overline{A}$.

Negating this equivalence gives the neighbourhood characterization