Metric Space is Hausdorff/Proof 2

Proof
$M$ is not Hausdorff.

That is, there are $x, y \in A: x \ne y$ such that:


 * $\forall \epsilon \in \R_{>0}: \exists z \in \map {B_\epsilon} x \cap \map {B_\epsilon} y$

where $\map {B_\epsilon} x$ denote the open $\epsilon$-ball of $x$ in $M$.

Let $r = \dfrac {\map d {x, y} } 2$.

Let $z \in \map {B_r} x \cap \map {B_r} y$.

Then:

This contradicts.

Thus, $M$ has to be Hausdorff.