In an arbitrary metric space, does it follow that if $x \ne y$ and $\epsilon \neq \delta$ then $B_\epsilon(x) \ne B_\delta(y)$?
I would say this is not true. We can find a counterexample in the $(X, d_0)$, the discrete metric space. If we take $B_\epsilon(x)$ and $B_\delta(y)$ such that $x \ne y$ and $\epsilon \neq \delta$ with both $\epsilon$ and $\delta$ greater than $1$, then these open balls will be equal as they will both contain all points of $X$. Have I got that right?