The problem is attached as a picture.
The Attempt at a Solution
I have been trying a lot to prove this without any really fruitful approach. At first I thought that the statement was false, or that you could at least construct a sequence of rationals such that V=(0,1) the following way:
Let a1 be a real in (0,1). Since the rationals are dense there exists a rational number b1 such that d(a1,b1)<ε. Let this be the first rational number in the sequence. Now let a2 be another real. Because the rationals are dense there exists a rational b2 such that d(a2,b2)<ε/2 etc. etc. and by successive use of this method I could generate the whole (0,1) with my approach. But this fails because (0,1) is not countable. So I'm open for any other approach to this problem.