TITLE: If $X$ is a topological separable space then every open set is a union of a countable number of open balls?
QUESTION [8 upvotes]: I am studying some real analysis (I am in the part before defining
Borel's $\sigma$-algebra) and there is a reminder in my lecture notes
that states that if $X$ is a separable metric space then every open
set is a union of a countable number of open balls (where separable
is defined as having a countable dense subset).
Can we remove the requirement that $X$ is metric ? 
To be honest,
I don't really know the proof even with requiring that $X$ is metric,
I guess that we move from any union to a countable union by using
somehow the fact that every open ball have an element in the dense
set, this is what also led me to believe that the claim doesn't have
much to do with $X$ being metric.
So there are two questions here, is it true for any topological space
and how can we prove this claim ?

REPLY [2 votes]: Berci has pointed out in a comment that, in order to talk about open balls, one needs a metric; there is no notion of ball in a general topological space.  It might also be worth pointing out that, in separable metric spaces, one has the stronger property that there is a countable family of open balls (namely those with rational radii and centers in a given countable dense set) such that every open set is a union of a subfamily of this particular family.  One could try to generalize this last observation by asking whether, in a separable topological space, there must be a countable family $\mathcal B$ of open sets such that every open set is the union of a subfamily of $\mathcal B$.  This property of a space is called the second axiom of countability (and such spaces are called second-countable), and it is known not to follow from separability.  Perhaps the nicest counterexample is the product (with the usual product topology) of $\kappa$ copies of a 2-element discrete space.  This is separable for all cardinal numbers $\kappa$ up to and including the cardinality of the continuum (and not for any larger $\kappa$), but it is second-countable only for countable $\kappa$.  
Some tangential comments: Instead of generalizing the observation that I began with, one could similarly try to generalize the original, weaker statement: Must there be a family $\mathcal B$ of open sets such that every open set is a union of a subfamily?  Unfortunately, this generalization is silly; the answer is trivially affirmative (whether or not the space is separable) as we can just take $\mathcal B$ to contain all the open sets.
Although I described what I consider the "nicest" counterexample above, I must admit that my personal favorite is another counterexample: the Stone-Cech compactification of a countably infinite discrete space.
Finally, I emphasize that, although I'm now a senior citizen, terrible terminology like "first-" and "second-countable", "first" and "second category", was around long before I learned it.  I officially disclaim any responsibility.