Here is Prob. 1, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:
Prove that if $X$ is an ordered set in which every closed interval is compact, then $X$ has the least upper bound property.
My Attempt:
Let $A$ be any non-empty subset of $X$ such that $A$ is bounded above in $X$; let $U$ be the set of all the upper bounds in $X$ of this set $A$. Then $U$ is also a non-empty subset of $X$.
Case 1.
If $A \cap U \not= \emptyset$, let us suppose that $v \in A \cap U$.
Now as $v \in U$, so $v$ is an upper bound of the set $A$, by our definition of the set $U$.
Also, as every element of $U$ is an upper bound of the set $A$ and as $v \in A$, so we must have $$ v \leq u \ \mbox{ for every element } u \in U. $$
Thus $v$ is an upper bound of the set $A$ and $v \leq u$ for every upper bound $u$ of $A$. Therefore $v$ is the least upper bound of the set $A$.
Case 2.
Let us now suppose that $A \cap U = \emptyset$.
Then for every element $a \in A$ and for every element $u \in U$, we must have $a < u$.
Let $a \in A$ and $u \in U$ be arbitrary. Then $a < u$. Moreover, the closed interval $[a,u]$ is compact, by our hypothesis.
Let $\mathscr{L}$ be the collection of all the closed intervals of the form $[x, y]$, where $x \in A$, $y \in U$, and $[x,y] \subset [a,u]$. That is, let $$ \mathscr{L} \colon= \big\{ \ [x, y] \ \colon \ x \in A, y \in U, a \leq x < y \leq u \ \big\}. \tag{*} $$ Then the interval $[a, u]$ itself is in $\mathscr{L}$ so that $\mathscr{L}$ is non-empty.
If $\left[x_1, y_1 \right], \ldots, \left[x_n, y_n \right]$ be any finite subcollection of $\mathscr{L}$, then we have $$ \bigcap_{j=1}^n \left[ x_j, y_j \right] = \left[ x_0, y_0 \right], $$ where $$ x_0 \colon= \max \left\{ x_1, \ldots, x_n \right\}, \qquad \mbox{ and } \qquad y_0 \colon= \min \left\{ y_1, \ldots, y_n \right\}. $$ Thus $\bigcap_{j=1}^n \left[ x_j, y_j \right]$ is again an interval in $\mathscr{L}$ and is therefore non-empty.
Then $\mathscr{L}$ is a collection of non-empty closed sets in $[a,u]$ such that $\mathscr{L}$ has the finite intersection property. Since $[a,u]$ is compact, therefore the intersection of all the closed intervals in $\mathscr{L}$ is non-empty, by Theorem 26.9 in Munkres.
That is, there is an element $p \in [a,u]$ such that $p \in [x,y]$ for all $x \in A$ and for all $y \in U$ for which $[x,y] \subset [a,u]$. In particular, this $p \in [a, u]$ also, so that $$ a \leq p \leq u. \tag{1} $$
But $a$ was chosen to be an arbitrary element of set $A$, and $u$ was chosen to be an arbitrary element of the set $U$ (i.e. $u$ was chosen to be an arbitrary upper bound for the set $A$).
Thus the leftmost inequality in (1) implies that $p$ is an upper bound of the set $A$, and the rightmost inequality in (1) implies that $p$ is also the least of all of the upper bounds of $A$.
Hence $p$ is the least upper bound of the set $A$.
But as $A$ was chosen to be an arbitrary non-empty subset of $X$ which was bounded above in $X$, so we can conclude that $X$ has the least upper bound property.
Is this proof correct? If so, is it clear enough? If not, then where are the issues?
