Closed Intervals form Neighborhood Basis in Real Number Line

Theorem
Let $\R$ be the real number line considered as a metric space under the usual metric.

Let $a \in \R$ be a point in $\R$.

Let $\mathcal B_a$ be defined as:
 * $\mathcal B_a := \left\{ {\left[{a - \epsilon \,.\,.\, a + \epsilon}\right]: \epsilon \in \R_{>0} }\right\}$

that is, the set of all closed intervals of $\R$ with $a$ as a midpoint.

Then the $\mathcal B_a$ is a basis for the neighborhood system of $a$.

Proof
Let $N$ be a neighborhood of $a$ in $M$.

Then by definition:
 * $\exists \epsilon' \in \R_{>0}: B_\epsilon' \left({a}\right) \subseteq N$

where $B_\epsilon' \left({a}\right)$ is the open $\epsilon'$-ball at $a$.

From Open Ball in Real Number Line is Open Interval:
 * $B_\epsilon' \left({a}\right) = \left({a - \epsilon \,.\,.\, a + \epsilon}\right)$

Let $\epsilon \in \R_{>0}: \epsilon < \epsilon'$.

Then by definition of closed interval and open interval:
 * $\left[{a - \epsilon \,.\,.\, a + \epsilon}\right] \subseteq \left({a - \epsilon' \,.\,.\, a + \epsilon'}\right)$

From Subset Relation is Transitive:
 * $\left[{a - \epsilon \,.\,.\, a + \epsilon}\right] \subseteq N$

Also by definition of closed interval and open interval:
 * $\left({a - \epsilon \,.\,.\, a + \epsilon}\right) \subseteq \left[{a - \epsilon \,.\,.\, a + \epsilon}\right]$

From Open Real Interval is Open Ball, $\left({a - \epsilon \,.\,.\, a + \epsilon}\right)$ is the open $\epsilon$-ball at $a$.

Thus by definition, $\left[{a - \epsilon \,.\,.\, a + \epsilon}\right]$ is a neighborhood of $a$ in $M$.

Hence there exists a neighborhood $\left[{a - \epsilon \,.\,.\, a + \epsilon}\right]$ of $a$ which is a subset of $N$.

Hence the result by definition of basis for the neighborhood system of $a$.