= Solution
Suppose $m$ is the <greatest lower bound> of the <set> $X$ of all <upper bounds> of $S$. Every $s\in S$ satisfies $s\preceq x$ for every $x\in X$, so each such $s$ is a <lower bound> of $X$. Since $m$ is the greatest of those <lower bounds>, $s\preceq m$ 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 $m\preceq x$ for every $x\in X$. It is therefore an <upper bound> of $S$ below every other <upper bound>, which proves \b[$m=\operatorname{lub}(S)$]. The argument also covers empty $S$ or $X$ whenever the assumed <greatest lower bound> exists.
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