I don't think your counterexample holds up, because I don't think I think At least as I understand it, the set of boundary points of a set are all those points which, if you take any neighborhood around one of them, that neighborhood always contains points of S and the compliment of S. The irrationals work for this definition just as well as the rationals. Thus, you get the entire interval.

So how would you go about proving your main result?