Let $A\subseteq X$ be nonempty and bounded above by $z$ . Let $a\in A$ . $[a,z]$ is closed and compact in the order topology (which is the same at the subspace topology, since the closed interval is convex). Consider the collection $\mathcal{L}$ of all intervals $[a',z']\subseteq[a,z]$ such that $a'\in A$ and $z'$ is an upper bound of $A$ . $\mathcal{L}$ is nonempty as $[a,z]\in\mathcal{L}$ . None of those intervals are empty, and they are all closed in $[a,z]$ . Moreover, the intersection of any finite number of intervals in $\mathcal{L}$ is also an interval in $\mathcal{L}$ and, therefore, is a nonempty set. Given that $[a,z]$ is compact, the intersection of all intervals in $\mathcal{L}$ has a point $x$ . Also, we have $x\ge a'$ for all $a'\in A$ (in $[a,z]$ or below) and $x\le z'$ for all upper bounds $z'$ of $A$ (in $[a,z]$ or above). There cannot be two such points $x<x'$ , as in this case both have to upper bounds of $A$ but then $x'\notin[a,x]$ . Therefore, there is a unique point $x$ in the intersection which is the least upper bound of $A$ .