So suppose that all the sets are proper subsets of X. Choose x_0 in X. Since a compact metric space is bounded, some ball centred at x_0, with sufficiently large radius R, will be the whole of X (and thus will not be a subset of any of the s). Therefore a Lebesgue number of the cover must be less that R. Hence the set of all Lebesgue numbers of the cover is bounded above. You have already proved that it is nonempty. Therefore it has a supremum, call it δ. We want to show that δ is the largest Lebesgue number of the cover.
The problem is that the supremum of a set need not always belong to the set. So we need to prove that δ is a Lebesgue number of the cover. To do this, we have to show that every ball of radius less than δ is contained in some member of the cover. So let B(x,r) be a ball centred at some point x, with radius r<δ. By the definition of supremum, there exists a number ε with r<ε<δ, such that ε is a Lebesgue number of the cover. But since r<ε it follows from the definition of Lebesgue number that any ball of radius r (in particular B(x,r)) must lie in some , which is what we wanted to prove.
[I'm sure that if Henri Lebesgue were still alive, he would want to point out that his name is spelt with a g, not a q.]