Solution
= 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
$$
\boxed{x\in\overline A\iff U\cap A\neq\varnothing\text{ for every open }U\ni x.}
$$