Show that every open ball is unique I need to show that if $B_r(x)$ = $B_s(y)$, this implies that $r=s$ and $x=y$. 
I also need to determine whether this holds only in $\mathbb{R}^n$ or in a more general metric space also. 
Cheers,
 A: What you want to prove is false in general. For example, there exist metric spaces for which every point of a given ball is a center of the ball.
A: It's true in any normed space.
Suppose that $x\neq y$.
Assume without loss of generality that $r\geq s$. Let $\displaystyle \epsilon=\frac {\min(||y-x||, r)}2$
Note that $\displaystyle x-(r-\epsilon)\frac{y-x}{||y-x||}\in B_r(x)=B_s(y)$.
Hence $\displaystyle \left|\left|y-\left(x-(r-\epsilon)\frac{y-x}{||y-x||}\right)\right|\right|\leq s$, equivalently $||y-x||+r-\epsilon\leq s$. Given the definition of $\epsilon$, we get  $r<s$ which contradicts $r\geq s$.
Therefore $x=y$. The equality $r=s$ follows easily.
A: Hint: Given a ball you can try to get your hands on $x$ and $r$. For example, there is going to be a unique point inside the ball which is furthest from its complement (which means, the infimum of the distances over all points on the boundary). Compactness (of the closure of the ball) will tell you existence of a point that maximizes this distance function, then you can argue for uniqueness. (Assume there are two points of maximal distance...) 
Probably there is a slicker way to organize this, but this kind of idea will at least get you started.
For the counter example: Consider the metric space where $d(x,y) = 0$ if $x =y $ and $d(x,y) =1 $ otherwise.
A: 
Lemma. For any closed ball $B=\{x:|x-x_0|\leq r\} \subset \mathbb R^n$ there exist $x_1,y_1$ such that 
  $$
|x_1-y_1|=\max_{x,y \in B}|x-y|=2r
$$
  and $x_0=\frac{x_1+y_1}2$.

Proof: By triangle inequaliry $|x-y|\leq |x-x_0|+|y-x_0|\leq 2r$ and equality holds only if the triangle $xyx_0$ degenerates to a segment $xy$ that include $x_0$ (this is also true in every strictly convex normed space). Since $B \subset \mathbb R^n$ is compact (this can be false for infinite dimensional normed spaces) the maximum of $|x-y|$ is achieved, so we have a segment $x_1y_1$ containing $x_0$ with $|x_1-y_1|=2r$ and since $|x_1-x_0|\leq r$, $|y_1-x_0|\leq r$ and $|x_1-x_0|+|y_1-x_0|= 2r$ we conclude $|x_1-x_0|=|y_1-y_0|=r$.

Corollary. If $B=B_r(x)=B_s(y)\subset \mathbb R^n$ then $r=s$ and $x=y$

Proof: If we take $x_1,y_1 \in \overline{B}$ from the lemma we must have $r=2|x_1-y_1|=s$ and $x=\frac{x_1+y_1}2=y$.
