In Rudin's Principles of Mathematical Analysis, we have the following theorem early on:
Theorem 1.11. 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 $\alpha = \inf B$.
In particular, $\inf B$ exists in $S$.
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 (see Definition 1.8) $\gamma$ is not an upper bound of $L$, hence $\gamma \not\in B$. It follows that $\alpha \leq x$ for every $x \in B$. Thus, $\alpha \in L$.
If $\alpha < \beta$ then $\beta \not\in L$, since $\alpha$ is an upper bound of $L$.
We have shown that $\alpha \in L$ but $\beta \not\in L$ if $\beta > \alpha$. In other words, $\alpha$ is a lower bound of $B$, but $\beta$ is not if $\beta > \alpha$. This means that $\alpha = \inf B$.
Why it important that $S$ has the least upper bound property? Going through the proof, I don't see what would change if $S$ didn't have the LUBP.
Additionally, Rudin says beforehand that LUBP implies GLBP for an arbitrary set, but all I see is that if some set $A$ in a set $S$ has LUBP, there exists another set $B$ in $S$ that has GLBP. Is that the result we wanted?