Closure of Intersection may not equal Intersection of Closures

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H_1$ and $H_2$ be subsets of $S$.

Let $H_1^-$ and $H_2^-$ denote the closures of $H_1$ and $H_2$ respectively.

Then it is not necessarily the case that:
 * $\left({H_1 \cap H_2}\right)^- = H_1^- \cap H_2^-$

Proof
Note that from Closure of Intersection Subset of Intersection of Closures it is always the case that:
 * $\left({H_1 \cap H_2}\right)^- \subseteq H_1^- \cap H_2^-$

It remains to be shown that it does not always happen that:
 * $\left({H_1 \cap H_2}\right)^- = H_1^- \cap H_2^-$

Proof by Counterexample:

Let $\left({\R, \tau_d}\right)$ be the real number line under the usual (Euclidean) topology.

Let $H_1 = \left({0 \,.\,.\, \dfrac 1 2}\right)$ and $H_2 = \left({\dfrac 1 2 \,.\,.\, 1}\right)$.

By inspection it can be seen that:
 * $H_1 \cap H_2 = \varnothing$

Thus from Closure of Empty Set is Empty Set:
 * $\left({H_1 \cap H_2}\right)^- = \varnothing$

From Closure of Open Real Interval is Closed Real Interval:
 * $H_1 = \left[{0 \,.\,.\, \dfrac 1 2}\right], H_2 = \left[{\dfrac 1 2 \,.\,.\, 1}\right]$

Thus:
 * $H_1^- \cap H_2^- = \left\{{\dfrac 1 2}\right\}$

So $\left({H_1 \cap H_2}\right)^- \ne H_1^- \cap H_2^-$

Hence the result.