 Solution

Suppose $m$ is the greatest lower bound of the set $X$ of all upper bounds of $S$. Every $s\in S$ satisfies $s⪯x$ for every $x\in X$, so each such $s$ is a lower bound of $X$. Since $m$ is the greatest of those lower bounds, $s⪯m$ for every $s\in S$. Thus $m$ itself is an upper bound of $S$, meaning $m\in X$.

Also $m$, being a lower bound of $X$, satisfies $m⪯x$ for every $x\in X$. It is therefore an upper bound of $S$ below every other upper bound, which proves $m=\mathrm{lub}(S)$. The argument also covers empty $S$ or $X$ whenever the assumed greatest lower bound exists.