Any necessary and sufficient condition(s) for closure of an open ball to be the corresponding closed ball? Let $(X,d)$ be a metric space, $a \in X$, and $\delta$ be a positive real number. Then the open ball $B(a;\delta)$ is defined as 
$$B(a;\delta) \colon= \left\{ \ x \in X \ \colon \ d(x,a) < \delta \ \right\},$$
whereas the sphere $S(a; \delta)$ is defined as 
$$S(a;\delta) \colon= \left\{ \ x \in X \ \colon \ d(x,a) = \delta \ \right\},$$
Then the closure $\overline{B(a;\delta)}$ of $B(a;\delta)$ need not equal $B(a;\delta) \cup S(a;\delta)$. 
In particular, in the Euclidean space $\mathbb{R}^k$, this holds, whereas in a discrete metric space (with more than one point) this fails. Am I right? 
Now is (are) there any necessary and / or sufficient condition(s) on $(X,d)$ under which 
$$\overline{B(a;\delta)} = B(a;\delta) \cup S(a;\delta)?$$
 A: This is a global condition - that is, it is both necessary and sufficient to have your condition be true for all $x,\delta$.
You need:

(Condition 1): Given any $x\neq y$ and any $\epsilon>0$ that there is some $z$  so that $d(y,z)<\epsilon$ and$d(x,z)<d(x,y)$.

That is, every neighborhood of $y\neq x$ has a point closer to $x$ than $y$ is.
For example, the discrete space fails because for some $\epsilon>0$ there is no $z\neq y$ with $d(y,z)<\epsilon$.
It's necessary because if you have a counter-example to my condition, with and $x\neq y\in X$, define $r=d(x,y)$. Then $y$ is on the sphere or radius $r$ but, for some $\epsilon>0$, $B_{\epsilon}(y)\cap B_r(x)=\emptyset$, so $y$ is not in the closure of $B_r(x)$. 
It is sufficient because if $y$ is on the sphere of radius $r$ around $x$, then $d(x,y)=r$. Now, for each $\epsilon_k=\frac{1}{k}$, find  $z_k\in B_{\epsilon_k}(y)$ with $d(x,z_k)<d(x,y)=r$. Then $z_k$ is a sequence in $B_r(x)$ which converges to $y$, so $y$ is in the closure of $B_r(x)$.
This can be rewritten as:

(Condition 2): If $x\in X$ and $U$ is an open set not containing $x$, then the function $U\to\mathbb R^{+}$ defined as $u\mapsto d(x,u)$ does not achieve its infimum in $U$.

The relationship to condition (1) is more obvious, I suppose, if you rewrite condition 2 as:

(Condition 1.5): Given open $U$ and $x\notin U$, then for any $y\in U$, there is a $z\in U$ so that $d(x,z)<d(x,y)$. 

That's therefore clearly an extended version of Condition (1), applied to all open sets containing $y$, rather than just open balls around $y$.
Proof that Condition 1 and Condition 2 are equivalent
Assume Condition (1).
Let $U\subseteq X$ and $x\notin U$. 
For any $y\in U$, pick $\epsilon>0$ so that $B_{\epsilon}(y)\subseteq U$. This can be done because $U$ is open.
But condition (1) means that there must be a $z\in B_{\epsilon}(y)\subseteq U$ so that $d(x,z)<d(x,y)$. So $d(x,y)$ is not a lower bound for $\{d(x,u)\mid u\in U\}$, for any $y\in U$, proving condition $2$.
Assuming Condition (2):
Given $x\neq y\in X$. If $\epsilon>0$ is chosen, define $U=B_{\epsilon}(y)$. 
If $x\in U$, then $d(x,x)=0<d(x,y)$, so we can just choose $z=x$.
If $x\notin U$, then, since $U$ is open, we know by condition (2) that $d(x,y) \neq \inf_{u\in U} d(x,u)$, so there must be a $z\in U=B_{\epsilon}(y)$ with $d(x,z)<d(x,y)$.
Thus we have Condition (1).
A: An example of a meaningful, sufficient condition for this is that $X$ is a length space. This is certainly not a necessary condition: for example, a dense subspace of a length space also has this property (more generally, the property is inherited by dense subspaces; it is also inherited by open subspaces).
A simple necessary condition is that for each $x$, you need to know that within the range of the function $d(x,\cdot)\colon X\to {\bf R}$, every point is the limit of an increasing sequence. Otherwise, it is easy to find a counterexample. Of course, this condition is certainly not sufficient, as one can easily alter a space to artificially satisfy it (say, by taking a product with ${\bf R}$).
Another example of a space with this property is the (Euclidean) plane with a single open disk removed. This space is very far from being a length space, but it has your property. On the other hand, a plane without an open rectangle will not have the property. This shows that the condition is very sensitive with respect to the geometry of the space (considering how similar the two examples are -- you could even round off the corners of the rectangle to make them even more alike).
