It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal
to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric
$$
d(x,y)=
\begin{cases}
0\qquad&\text{if and only if $x=y$}\\
1&\text{otherwise}
\end{cases}
$$
The open unit ball of radius $1$ around any point $x$ is the singleton set $\{x\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything. 

I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal? 

Here is a characterization that is straight from the definitions, but which it seems may be useful when verifying that a particular space has the property.
For any metric space $(X,d)$, the following are equivalent:

For any $x\in X$ and radius $r$, the closure of the open ball of radius $r$ around $x$ is the closed ball of radius $r$.
For any two distinct points $x,y$ in the space and any positive $\epsilon$, there is a point $z$ within $\epsilon$ of $y$, and closer to $x$ than $y$ is.
That is, for every $x\neq y$ and $\epsilon\gt 0$, there is $z$ with $d(z,y)<\epsilon$ and $d(x,z)<d(x,y)$.

Proof. If the closed ball property holds, then fix any $x,y$ with $r=d(x,y)$. Since the closure of $B_r(x)$ includes $y$, the second property follows. Conversely, if the second property holds, then if $r=d(x,y)$, then the property ensures that $y$ is in the closure of $B_r(x)$, and so the closure of the open ball includes the closed ball (and it is easy to see it does not include anything more than this, since if $g$ belongs to the closure of $B_r(x)$ then $d(x,g) \le r$ and so $g$ must also belong to the closed ball of radius $r$ centered at $x$).
QED
