Interior of Union is not necessarily Union of Interiors

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

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

Let $H_1^\circ$ and $H_2^\circ$ denote the interiors of $H_1$ and $H_2$ respectively.

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

Proof
From Union of Interiors is Subset of Interior of Union:
 * $\left({H_1 \cup H_2}\right)^\circ \subseteq H_1^\circ \cup H_2^\circ$

It remains to be shown that it is not necessarily the case that:
 * $\left({H_1 \cup H_2}\right)^\circ = H_1^\circ \cup H_2^\circ$

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 $H_2 = \left[{\dfrac 1 2 \,.\,.\, 1}\right]$.

Then:

and:

Hence the result.

Also see

 * Closure of Intersection may not equal Intersection of Closures