Closed Real Interval is Neighborhood Except at Endpoints

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

Let $\closedint a b \subset \R$ be a closed interval of $\R$.

Then $\closedint a b$ is a neighborhood of all of its points except $a$ and $b$.

Proof
Let $a < c < b$.

Let $\epsilon < \map \min {b - c, c - a}$.

From the definition of positive it follows that $\epsilon \in \R_{>0}$.

Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$.

We have that $c + \epsilon < b$ and $a < c - \epsilon$.

Thus:
 * $\map {B_\epsilon} c \subseteq \closedint a b$

It follows that, by definition, $\closedint a b$ is a neighborhood of $c$.

Now consider $a$.

Let $\epsilon \in \R_{>0}$.

We have that $a - \epsilon < a$ and so $\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ does not lie entirely in $\closedint a b$.

Similarly, consider $b$.

Let $\epsilon \in \R_{>0}$.

We have that $b < b + \epsilon$ and so $\map {B_\epsilon} b = \openint {b - \epsilon} {b + \epsilon}$ does not lie entirely in $\closedint a b$.

It follows that, by definition, $\closedint a b$ is a neighborhood of neither $a$ nor $b$.