Two different points of a metric space are contained in disjoint open balls To prove: Let $(X,d)$ be a metric space and $x,y \in X$ two different points. Show that there exist $r_x, r_y \gt 0$ such that $B(x, r_x) \cap B(y,r_y) = \varnothing$.
My solution: Let $r_x, r_y = {d(x,y)\over 2}$. Suppose that $B(x, r_x) \cap B(y,r_y) \neq \varnothing$. This means that there exists $z \in X$ such that $z \in B(x, r_x) \cap B(y,r_y)$. By definition, $d(x,z)<r_x$ and $d(z,y)<r_y$. 
Using the triangular inequality, we have:
$d(x,y) \le d(x,z)+d(z,y)<r_x+r_y={d(x,y)\over 2}+{d(x,y)\over 2}=d(x,y)$, so we get $d(x,y)<d(x,y)$, which is a contradiction.
Is my proof OK? Is there another proof not by contradiction? Thanks.

Your proof is fine.
You could slightly reword your proof to make it appear in natural language to not be a proof by contradiction. I don't think this fundamentally changes anything, but it might make the proof read a little more nicely.

Let $z$ be arbitary, and let us show that $z \notin B(x,r_x) \cap B(y,r_y)$. Without loss of generality, assume that $d(x,z) \geq d(y,z)$. Then
  $$
2r_x = d(x,y) \leq d(x,z) + d(y,z) \leq 2d(x,z).
$$
  Hence $d(x,z) \geq r_x$, so $z \notin B(x,r_x)$, as we wanted to show.

