(1) Show that implies , and
(2) Show that if is an upper bound, then .
For the first one, pick any , then . Since we know because , we can safely divide by to get the inequality .
So we know that 1 is an upper bound of S. The next step is to prove that there are no upper bounds of the set that are less than 1.
First it should be clear that everything in S is positive. Negative numbers are never larger than their square, which are always positive, and we showed 0 is also not a member.
Suppose for contradiction that 1 were not the Sup. Then there is some which is an upper bound of S, but also so . First note that is a positive number because everything in S is positive, and is an upper bound of the set. So we can multiply the inequality by without changing the sign. So implies . So it follows that .
Now consider the number , which is the number lying halfway between and 1. Now you can easily prove that its square is smaller than itself -- that is, , which means that creature is also in S.
But here's the contradiction. We said was an upper bound, which means it's greater-than-or-equal-to everything in S. But we found an element, which lies between and 1, that lies in S but is bigger than . Therefore, no such exists, so 1 must be the least upper bound, i.e., the Sup of S.