Closure of Real Interval is Closed Real Interval/Proof 2

Proof
Let $I$ be one of the intervals as specified in the exposition.

Note that:
 * $(1): \quad$ By Condition for Point being in Closure, $x \in I^-$ every open set in $\R$ containing $x$ contains a point in $I$.


 * $(2): \quad$ From Union of Open Sets of Metric Space is Open, every open set in $\R$ is a union of open intervals.

Thus we also have that $x \in I^-$ every open interval containing $x$ also contains a point in $I$.

This equivalence will be made use of throughout.

Lemma 2
By the two lemmas proven above:
 * $\closedint a b = I^-$