Half-Open Real Interval is Closed in some Open Intervals

Theorem
Let $\R$ be the real number line considered as an Euclidean space.

Let $\hointr a b \subset \R$ be a half-open interval of $\R$.

Let $c < a$.

Then $\hointr a b$ is a closed set of $\openint c b$.

Similarly, let $d > b$.

Then the half-open interval $\hointl a b$ is a closed set of $\openint a d$.

Proof
Consider:
 * $A := \openint c b \setminus \hointr a b = \openint c a$

Then $A$ is an open interval.

By Open Real Interval is Open Set, $A$ is open in $\R$.

By the definition of the subspace topology it follows that $A$ is open in $\openint c b$.

Thus by the definition of closed set, $\openint c b \setminus A = \hointr a b$ is closed in $\openint c b$.

Mutatis mutandis, the argument also shows that $\hointl a b \subset \R$ is a closed set of $\openint a d$.