Separability almost everywhere Let $(X,d)$ be a metric space and $\mu$ a Borel probability measure. 
Suppose that for every $\epsilon>0$ we have that $\mu(B_{\epsilon
}(x))=c_{\epsilon}>0$ a.e. 
Is this enough to show that for every $\epsilon$ there exist  a countable set $K\subset X,$ such that
$\mu(\cup_{x\in K}B_{\epsilon}(x))=1?$
 A: Your assumption implies that your metric space is separable.
Exercise: Suppose a metric space is not separable. Then for some $\epsilon > 0$,  one can find an uncountable family of pairwise disjoint open balls each of radius $\epsilon$.
So you can take $K$ to be a countable dense subset of $X$.
A: As shown by user 87216, the answer is "yes". 
Here is an answer to a question that has not been asked; namely, does the conclusion hold true ... without any assumption on $\mu$? Roughly speaking, the answer is that if the topology of $X$ has a basis which is "not too big", then the answer is "yes".
Precisely, a cardinal $\kappa$ is said to have measure $0$ if there is no non-zero finite measure defined on all subsets of a set with cardinality $\kappa$ for which every point has measure $0$. Cardinals which do not have measure $0$ are necessarily "very large" (but I'm not able to say much more).
In Oxtoby's book Measure and category, one can find the following result: if $Z$ is a metric space with a basis whose cardinal has measure $0$, and if $\nu$ is a finite Borel measure on $Z$, then the union of any family of open sets with $\nu$-measure $0$ also has $\nu$-measure $0$. (This is Theorem 16.4).
Now, assume that your metric space $(X,d)$ has a basis whose cardinal has measure $0$, and let us try to show that the answer to your question is "yes" in this case.
Let $\mu$ be an arbitrary Borel probability measure on $X$. Fix $\varepsilon >0$, and denote by $\mathcal A$ the family of all sets $A\subset X$ which are unions of countably many balls of radius $\varepsilon$. We have to find a set $A\in\mathcal A$ such that $\mu (A)=1$.
Put $c=\sup\{ \mu(A);\; A\in\mathcal A\}$. Then one can find a sequence $(A_n)\subset \mathcal A\}$ such that $\mu(A_n)\to c$, and the set $A=\bigcup_n A_n$ is in $\mathcal A$ and satisfies $\mu (A)=c$. It is enough to show that $\mu(X\setminus A)=0$.
Put $Z:=X\setminus A$. Then, by the very definition of $Z$, the following holds true: for any $z\in Z$ we have $\mu(B_\varepsilon(z)\setminus A)=0$. Indeed, if $\mu(B_\varepsilon(z)\setminus A)>0$ for some $z$, then the set $A'=A\cup B_\varepsilon (z)$ is in $\mathcal A$ an satisfies $\mu(A')>c$, which contradicts the definition of $c$.
Now, consider the metric space $(Z,d)$ and the measure $\nu=\mu_{|Z}$. The preceding paragraph says that any open ball with radius $\varepsilon$ in $Z$ has $\nu$-measure $0$. Since $Z$ satisfies the same topological assumption as $X$, it follows from the above theorem from Oxtoby's book that $\nu=0$, i.e. $\mu (Z)=0$ as required.
Without any assumption on the metric space $X$ and the measure $\mu$, the answer can be "no". Consider a set $X$ with cardinality $\kappa$, where $\kappa$ does not have measure $0$, put the discrete measure on $X$ and take any probability measure $\mu$ defined on all subsets of $X$ witnessing that $\kappa$ does not have measure $0$. Then all open balls with radius $1$ have measure $0$, so $X$ cannot be written as a countable almost everywhere union of such balls.
