Distinct Points in Metric Space have Disjoint Open Balls/Proof 2

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

Then $\map d {x, y} > 0$.

Put $\epsilon = \dfrac {\map d {x, y} } 2$.

Let $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ be the open $\epsilon$-balls of $x$ and $y$ respectively.

Suppose $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ are not disjoint.

Then $\exists z \in M$ such that $z \in \map {B_\epsilon} x$ and $z \in \map {B_\epsilon} y$.

Then $\map d {x, z} < \epsilon$ and $\map d {z, y} < \epsilon$.

Hence $\map d {x, z} + \map d {z, y} < 2 \epsilon = \map d {x, y}$.

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

Hence the open balls must be disjoint.