We need to show:
1)"a" is an upper bound.
2)"a" is least upper bound.
#1)It is an upper bound, for that is the definition of this non-empty.
#2)Say that "a" is not a least upper bound say b<a is the least upper bound. Then by the denseness of Q we have that there exists an rational "r" such that b<r<a hence, r in the set. But then "b" cannot be an unpper bound. A contradiction. Thus "a" is the least upper bound.