Distinct Points in Metric Space have Disjoint Open Balls

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

Let $$x, y \in M: x \ne y$$.

Then there exist disjoint $\epsilon$-neighborhoods $$N_\epsilon \left({x}\right)$$ and $$N_\epsilon \left({y}\right)$$ containing $$x$$ and $$y$$ respectively.

Proof
Let $$x, y \in A: x \ne y$$.

Then $$d \left({x, y}\right) > 0$$.

Put $$\epsilon = \frac {d \left({x, y}\right)} 2$$.

Let $$N_\epsilon \left({x}\right)$$ and $$N_\epsilon \left({y}\right)$$ be the $\epsilon$-neighborhoods of $$x$$ and $$y$$ respectively.

Suppose $$N_\epsilon \left({x}\right)$$ and $$N_\epsilon \left({y}\right)$$ are not disjoint.

Then $$\exists z \in M$$ such that $$z \in N_\epsilon \left({x}\right)$$ and $$z \in N_\epsilon \left({y}\right)$$.

Then $$d \left({x, z}\right) < \epsilon$$ and $$d \left({z, y}\right) < \epsilon$$.

Hence $$d \left({x, z}\right) + d \left({z, y}\right) < 2 \epsilon = d \left({x, y}\right)$$.

This contradicts the definition of a metric, so there can be no such $$z$$.

Hence the neighborhoods must be disjoint.