 Solution

The closure $\overline{A}$ is the intersection of every closed subset of $X$ containing $A$. Thus it is the smallest such closed subset. Equivalently,For equivalence, if an open neighbourhood misses $A$, its complement is a closed superset of $A$ missing $x$. Conversely, a closed superset missing $x$ has an open complement that is a neighbourhood of $x$ disjoint from $A$.