Union of Regular Open Sets is not necessarily Regular Open

Theorem
Let $T = \left({S, \tau}\right)$ 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:
 * $\left({0 \,.\,.\, \dfrac 1 2}\right), \left({\dfrac 1 2 \,.\,.\, 1}\right)$

are both regular open sets of $\R$.

Consider $A$, the union of the adjacent open intervals:
 * $A := \left({0 \,.\,.\, \dfrac 1 2}\right) \cup \left({\dfrac 1 2 \,.\,.\, 1}\right)$

From Interior of Closure of Interior of Union of Adjacent Open Intervals:
 * $A^{- \circ} = \left({0 \,.\,.\, 1}\right)$

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