Rudin Theorem $1.11$ After spending a few hours trying to understand Theorem $1.11$ in Rudin's Principles of Mathematical Analysis, I still don't follow the proof.

$1.11$ Theorem Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.

This is what I understand so far:
$B$ is bounded below means that $L$ is not empty and $L = \{ y \ | \ y \leq x \ \forall x \in B \}$. Then every $x \in B$ is an upper bound of $L$, which means that $L$ is bounded above. Since $L \subset S$, $L$ not empty, and $L$ is bounded above that implies that $\sup L = \alpha \in S$. And because $\alpha = \sup L$, $\gamma < \alpha$ implies that $\gamma$ is not an upper bound of $L$ and $\gamma \notin B$ since every element of $B$ is an upper bound of $L$. 
This is where I get confused:
Since $B$ is bounded below, there exists an $\omega \in S$ such that $\omega \leq x \ \forall x \in B$. Then Rudin claims, "It follows that $\alpha \leq x$ for every $x \in B$." Can someone explain why that is true or at least give me a hint?
Thanks in advance.
 A: Rewording it slightly: Every element of $B$ is an upper bound for $L$, so if $x\in B$ is less than $\alpha$, then $x$ is an upper bound for $L$ smaller than the least upper bound.  This contradicts the definition of least upper bound, so no such $x$ exist.  In other words, $\alpha\leq x$ for every $x\in B$.
I do not know why you included a restatement of the fact that $B$ is bounded below. The part you put in quotes follows directly from the last sentence of the previous paragraph in your post:

And because $\alpha = \sup L$, $\gamma < \alpha$ implies that $\gamma$ is not an upper bound of $L$ and $\gamma \notin B$ since every element of $B$ is an upper bound of $L$.

From here you could use contraposition:
$$\gamma<\alpha\implies \gamma\not\in B$$
is equivalent to 
$$\gamma\in B \implies \gamma\geq \alpha.$$
A: I think that Rudin has complicated the proof a little bit. Here is what I learned from his proof to write my own. I think this is more straight forward.
Let $L$ be the set of all lower bounds of $B$ or more precisely $L=\{l\,|\,\forall x:x\in B\implies x\ge l\}$. Since $B$ is bounded below so $L$ is not empty. Also, for every $x$ in $B$ and every $l$ in $L$ we have $l\le x$. This means that every $x\in B$ is an upper bound for $L$. By the L.U.B. property of $S$, we conclude that $L$ has a supremum denoted by $\sup L$. As $\sup L$ is smaller than any other upper bound of $L$ we conclude that $\sup L \le x$ for every $x\in B$. But this means that $\sup L$ is also a lower bound of $B$ and hence $\sup L \in L$. Furthermore, if $l\in L$ is any lower bound of $B$ then $l \le \sup L$. So far, we proved $\sup L$ is a lower bound of $B$ and is greater than any other lower bound of $B$ so it is the greatest lower bound of $B$, that is $\inf B =\sup L$. Accordingly, the set $S$ has the G.L.B. property.
