Can the corollary to the completeness axiom (the least upper bound property) be proved using the supremum of the set of lower bounds? The corollary in question is, of course, the one regarding the existence of $\inf S$ for some nonempty subset $S$ of $\mathbb{R}$, that is bounded below.
I am already aware of the proof that involves taking the set $-S:=\{-s:s\in S\}$ and showing that $\inf S = -\sup (-S)$. I'm wondering if you could argue a different proof however; that if the set of the lower bounds of $S$ is $L$, then $\inf S = \sup L$, thus proving the corollary. I know that $\sup L$ exists because of the completeness axiom, however I don't have a clue how to show that $\sup L \in L$, or if you even can show it. 
Would an argument such as this be sound? And if so, how would you go about showing that $\sup L \in L$?
I'm taking my first course in real analysis so I'm sorry if this is a stupid question.

Suppose $\sup L\notin L$. Then $\sup L$ is not a lower bound for $S$. Which means $\exists s\in S$ such that $\sup L>s$. Then $s$ cannot be an upper bound for $L$. Which means $\exists l\in L$ such that $l>s$; contradicting the fact that $l$ is a lower bound for the set $S$. So you get $\sup L\in L$. This shows that the set of lower bounds has a maximum element, viz. $\sup L$. It is a lower bound for S, which is greater than all the other lower bounds. Then by definition, it is $\inf S$.
