Closure of Real Interval is Closed Real Interval/Proof 2/Lemma 2

Lemma for Closure of Real Interval is Closed Real Interval
Let $I$ be a non-empty real interval such that one of these holds:
 * $I = \openint a b$
 * $I = \hointr a b$
 * $I = \hointl a b$
 * $I = \closedint a b$

Let $I^-$ denote the closure of $I$.

Then:
 * $x \notin \closedint a b \implies x \notin I^-$

Proof
Suppose $x \notin \notin \closedint a b$.

We must find an open interval containing $x$ which does not contain an elements in $I$.

There are two possibilities:
 * $x < a$

or:
 * $x > b$

Case: $x < a$
By Real Numbers are Densely Ordered, there exists $r \in \R$ such that $x < r < a$.

Thus $\openint {x - 1} r$ is an open interval, all of whose elements are less than $a$, and hence not in $I$.

Case: $x > b$
This is similarly to the case when $x < a$.

Here instead we pick $r$ such that $b < r < x$, and consider the open interval $\openint r {x + 1}$.