Question: Prove that every non-empty open set in is the disjoint union of a countable collection of open intervals.
The key things to prove are the disjointness and the countability of such open intervals. Otherwise, if disjointness and countability are not required, we may simply take a small open interval centered at each point in the open set, and their union will be the open set.
Elementary Proof: Let be a non-empty open set in .
Let . There exists an open interval containing . Let be the maximal open interval in containing , i.e. for any open interval containing , . (The existence of is guaranteed, we can take it to be the union of all open intervals containing .)
We note that such maximal intervals are equal or disjoint: Suppose and then is an open interval in containing , contradicting the maximality of .
Each of the maximal open intervals contain a rational number, thus we may write . Upon discarding the “repeated” intervals in the union above, we get that is the disjoint union of a countable collection of open intervals.
There are many other good proofs of this found here (http://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv), though some can be quite deep for this simple result.