Can this statement about supremum be proven directly? Theorem (I.32, pg 27, Apostol's Calculus Vol.1): Let $\varepsilon \in \mathbf{R}^+$ and let $S \subset \mathbf{R}$.  If $\sup S$ exists, then there exists $x\in S$ such that $$x > \sup S -\varepsilon.$$ 
(Reminder: The real number $B = \sup S$ is the least upper bound of $S$; that is, $B$ is an upper bound for $S$ and no number less than $B$ is an upper bound for $S$.)
Now this is proven very elegantly in 1 line, via contradiction.  For if $x\leq \sup S - \varepsilon$ for all $x\in S$ then $\sup S-\varepsilon$ would be an upper bound for $S$ smaller than its least upper bound.  I am wondering if this statement could be proven directly, with out arguing by contradiction.  
The reason I am interested in a direct proof is because I think I get a better, more permanent understanding of things when I hear their direct proof.  

For a direct prove: 
Let $\epsilon >0$  be given.  
Note that $ (\sup S -    \epsilon) < \sup S $ 
Therefore $ (\sup S -    \epsilon)$ is not an upper bound of S, because  $\sup S$ is the least upper bound of S. 
Thus there is an element of S, say $x$  such that $x> \sup S -    \epsilon$ 
