Quote by Dick
Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S.

?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.
Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.

In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.