Characterization of closed sets similar to open sets doesn't exist? In a topological space $X$, it is a simple result (using axiom of choice) that a subset $U$ is open if and only if each of $x\in U$ has an open neighborhood contained in $U$.
I was wondering if we can characterize the closed subsets in a similar manner, i.e., using closed neighborhoods of points of a closed set $K$. The above characterization uses the arbitrary union property of open sets and I reckon that the characterization that I am seeking should use the arbitrary (nonempty) intersection property of closed sets.
However, I am stuck. Any help?

Note that I am not looking for this characterization: $K$ is closed $\iff$ all points in $X\setminus K$ have an open neighborhood disjoint from $K$.

A set $K$ in a topological space $X$ is closed if and only if
$$
\bigcap_{\substack{K \subseteq C \\ C \text{ closed}}} C = K
$$
This uses the arbitrary intersection of closed sets that you wanted.  This just expresses that $K$ is closed if and only if $\overline{K}=K$.  (The left-hand side is just an alternate--and my favorite--definition of the closure of $K$.)  This is the exact dual of the following definition of the interior of a set $A$:
$$
\mathrm{int}A = \bigcup_{\substack{U \subseteq A \\ U \text{ open}}} U.
$$
And, of course, $A$ is open if and only if $\mathrm{int}A=A$.
Alternatively, use the fact that $K$ is closed if and only if it contains all its cluster points.  Thus $K$ is closed if and only if each point $x \in X$ that has the property that every open neighborhood of $x$ intersects $K$ necessarily has $x \in K$.
