In all the proofs I can find of the Open Mapping Theorem (for example here) at the outset it is mentioned that it is enough to prove that for all a in U, f(a) is contained in a disk that is itself contained in f(U).
How is that enough? One needs to prove that for every open set U' (that is a subset of U) the theorem holds, however the U used in that opening statement is a connected open set (and thus a stricter condition).
I find that every way I try and reconcile this I run into something non-trivial (e.g. Considering any open set a countable union of connected open sets. Can you do that in Rn? Does it require a finitely dimensioned space or any other such restriction?), even though the proof's statement suggests it is obvious that proving for U is enough. Stumped.
I'm aware that the statement proves that f(U) is open (as a union of open disks), that's not what I'm asking. My question is how is that enough to prove that f is open, i.e. f(V) is open for every V open subset of U (in particular since V may not be connected like U and certainly need not be a disk).