Closure of Open Ball in Metric Space

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

Let $$N_\epsilon \left({x}\right)$$ be an $\epsilon$-neighborhood in $$M = \left({A, d}\right)$$.

Let $$y \in \operatorname{cl} \left({N_\epsilon \left({x}\right)}\right)$$.

Then $$d \left({x, y}\right) \le \epsilon$$.

Proof
Suppose $$d \left({x, y}\right) > \epsilon$$.

Then $$N_{d \left({x, y}\right) - \epsilon} \left({y}\right)$$ is an open set containing $$y$$ and not meeting $$N_\epsilon \left({x}\right)$$.

Hence $$y \notin \operatorname{cl} \left({N_\epsilon \left({x}\right)}\right)$$.