Closure of Irrational Interval is Closed Real Interval

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

Let $\left({\R \setminus \Q, \tau_d}\right)$ 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:
 * $\Bbb I \cap \left({\R \setminus \Q}\right)$

is the closed real interval $\left[{a \,.\,.\, b}\right]$.

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

From Closure of Real Interval is Closed Real Interval:
 * $\Bbb I^- = \left[{a \,.\,.\, b}\right]$

From Closure of Irrational Numbers is Real Numbers:
 * $\left({\R \setminus \Q}\right)^- = \R$

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

From Intersection with Subset is Subset:
 * $\Bbb I^- \cap \left({\R \setminus \Q}\right)^- = \left[{a \,.\,.\, b}\right]$

and so:
 * $\left({\Bbb I \cap \left({\R \setminus \Q}\right)}\right)^- \subseteq \left[{a \,.\,.\, b}\right]$

From Irrationals are Everywhere Dense in Reals:
 * $\left[{a \,.\,.\, b}\right] \subseteq \left({\Bbb I \cap \left({\R \setminus \Q}\right)}\right)^-$

and the result follows.