Yes, this is a homework question so I'm looking for hints not complete work.
Prove the lemma: Let $S$ be a bounded and nonempty subset of $\Bbb{R}$, and let $\alpha = \sup(S)$. For all $\varepsilon>0$ then there is some $x \in S$ so that $\alpha-\varepsilon< x\leq \alpha$.
So my question is other than using the definition for $\sup(S)$ when $\alpha>0$, we get $x>\alpha-\varepsilon$. How would I prove that?
Notice that because $a=\sup S $, for $x\in S $, $x\le a $ is always true. So really we just want to show that for any $\epsilon > 0$ there is an $x\in S $ such that $a-\epsilon < x $. Suppose no such $x $ existed. Then $a-\epsilon $ is an upper bound of $S $, but $a-\epsilon < a $ and $a=\sup S $. See the contradiction? Also I'm not sure what you mean by "other that using the definition of $\sup $", because the definition of $\sup $ is crucial.