Proving that a countable set in an arbitrary set $ X $ has outer measure zero . It is well known that in $\mathbb{R} $ (well, also in $\mathbb{R}^n$) a countable set $A $ has outer measure zero. It is not really hard to prove and it is a common exercise among the books which I've already done.
However, I'm somewhat stuck at trying to prove it for an arbitrary set because I can't figure out how to let the $\varepsilon$-thing play into the game.
Is it still true that for every countable set $A\subseteq X$ one has that it's outer measure is zero ? And if so, which conditions one must have over $X$ ?
Thanks!

It depends on the (outer) measure and set $X$ you look at. For example let $X = \mathbb{Z}$ and define an outer measure that is not zero on each subset of $\mathbb{Z} $. Or on $\mathbb{R}$ define $\nu(\emptyset) := 0, \nu(A) := 1  \, (\emptyset \neq A \in \mathcal{P}(X))$. 
