In Rudin $1.11$ Theorem Proof he claims the following
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 $\alpha = \inf B$.
Proof. Since $B$ is bounded below, $L$ is not empty. Since $L$ consists of exactly those $y \in S$ which satisfy the inequality $y \leq x$ for every $x \in B$, we see that every $x \in B$ is an upper bound of $L$. Thus $L$ is bounded above. Our hypothesis about $S$ implies therefore that $L$ has a supremum in $S$ call it $\alpha$
If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $L$, hence $\gamma \notin B$. It follows that $\alpha \leq x $ for every $x \in B$. Thus $\alpha \in L$
If $\alpha < \beta$ then $\beta \notin L$ since $\alpha$ is an upper bound of $L$
We have shown that $\alpha \in L$ but $\beta \notin L$ if $\beta > \alpha$. In other words, $\alpha$ is a lower bound of $B$, but $\beta $ is not if $\beta > \alpha$. This means that $\alpha = \inf B$
I am confused in the following:
I don't follow why $L \subset S$ given $S$ is an ordered set with the least upper bound property and $B \subset S$, $B$ is not empty and $B$ is bounded below.
If $L$ is not a subset of $S$, then the assumption of the proof will not follow, I think I have missed something.
Can someone help me out? This proof is in Rudin's analysis page 5
Edit for clarification
Suppose the following let $S = (0, x]$, for which $x $ is some real positive number, we know $S$ is an ordered set with the least upper bound property, let $B = (0, y]$ for which $y < x$ and $y$ is positive real number, then $L = (-\infty, 0]$, we note that $\inf B = \sup L = 0$ however $0 \notin S$, thus we proved that an order set $S$ with the least upper bound property with $B = (0, y] \subset S \Rightarrow \inf B \notin S$