If 2 open balls define the same space, is it true that x=y and r=s? Let $(X,d)$ be a non-empty metric sapce, $r$ and $s$ are postive radii, and $b_r^{d}(x)=b^d_s(y)$ for some  $x,y \in X$. 
Is it true that $r=s$ ? 
Is it true that $x=y$?
My answer would be something like: no, because for example consider the discrete metric space $(X,d_0)$. Then $b_{35}(x)= X = b_{2}(y)$, where $2\neq 35$. Is this a good counter example?
The same would be true for $x \neq y$?
 A: Excellent counterexample. And as long as $X$ has more than $1$ element, taking any two balls of distinct radii $>1$ about distinct points will yield a double counterexample!
A: Let $X=[0,1]$. Then $B(0,2)=B(1,3)$.
Nota Bene: You can find some exotic counterexamples.
A: It's great you came up with those counterexamples. The discrete metric, and certain "cut out" subsets of some Euclidean space, make for great counterexamples to thoughts like that.
I'd just like to add that ultrametrics, and in particular $p$-adic metrics, can often serve as great (counter)examples for statements like that, and at the same time are not "artificial" in the sense of how the above might seem, and are one of the underlying tools of huge areas of research in pure mathematics, including Wiles' proof of Fermat's Last Theorem etc.
And once you've chosen your favourite prime number $p$, and look at the $p$-adic metric $d(x,y) = \lvert x-y \rvert_p$ (on $\mathbb Q$ or $\mathbb Z$, say), then e.g.
$$B_r(1) = B_s(y)$$
for all $y$ with $d(1,y) < 1$ and all radii $r$ and $s$ which lie in a common interval $(p^{n-1}, p^n)$ for $n \le 0$. Concretely, for $p=2$, e.g.
$$B_{0.7}(1) = B_{0.9}(7) = B_{8/13}(-17) = B_{5/9}(\frac{13}{189}) = B_{0.999}(\frac{-25}{7}) = ...$$
and, when meant as subsets of $\mathbb Q$, all these are a "metric" way to describe rationals with (when fully simplified) odd numerators; and, when restricted to $\mathbb Z$, these are just the odd numbers.
