Closure of Irrational Interval is Closed Real Interval

Theorem
Let $\struct {\R, \tau_d}$ be the real numbers under the usual (Euclidean) topology.

Let $\struct{\R \setminus \Q, \tau_d}$ be the irrational number space under the same topology.

Let $a, b \in \R$ such that $a < b$.

Let $\Bbb I \subseteq \R$ be an interval of $\R$

Then the closure of the set:
 * $\Bbb I \cap \paren {\R \setminus \Q}$

is the closed real interval $\closedint a b$.

Proof
Let $\Bbb I$ be an open real interval.

From Closure of Real Interval is Closed Real Interval:
 * $\Bbb I^- = \closedint a b$

From Closure of Irrational Numbers is Real Numbers:
 * $\paren {\R \setminus \Q}^- = \R$

From Closure of Intersection is Subset of Intersection of Closures:
 * $\paren{\Bbb I \cap \paren{\R \setminus \Q}}^- \subseteq \Bbb I^- \cap \paren {\R \setminus \Q}^-$

From Intersection with Subset is Subset:
 * $\Bbb I^- \cap \paren {\R \setminus \Q}^- = \closedint a b$

and so:
 * $\paren {\Bbb I \cap \paren {\R \setminus \Q} }^- \subseteq \closedint a b$

From Irrationals are Everywhere Dense in Topological Space of Reals:
 * $\closedint a b \subseteq \paren {\Bbb I \cap \paren {\R \setminus \Q} }^-$

and the result follows.