Union of Regular Open Sets is not necessarily Regular Open

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $U$ and $V$ be regular open sets of $T$.

Then $U \cup V$ is not also necessarily a regular open set of $T$.

Proof
Proof by Counterexample:

By Open Real Interval is Regular Open, the open real intervals:
 * $\openint 0 {\dfrac 1 2}, \openint {\dfrac 1 2} 1$

are both regular open sets of $\R$.

Consider $A$, the union of the adjacent open intervals:
 * $A := \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1$

From Interior of Closure of Interior of Union of Adjacent Open Intervals:
 * $A^{- \circ} = \openint 0 1$

Thus $A$ is not a regular open set of $\R$.