Munkres Chapter 27 Prob. 1 
Prove that if $X$ is an ordered set where every closed interval is compact, then X has a least upper bound.

Most solutions start off by considering closed intervals of the form $[a, z]$ where $a\in A \subset X$, and $z$ an upper bound of $A$. I am a little troubled by this. Should we assume that every ordered set has an upper bound in the first place? Or do we need to prove that too in this special case? The problem statement makes it very clear that there is a least upper bound, but the definition of least upper bound holds good only if there is an upper bound. Is that sufficient?
Please help me out.
 A: You misstated the problem.

The actual problem, as stated in the book, is this:


*
Prove that if $X$ is an ordered set in which every closed interval is compact, then $X$
has the least upper bound property.


Here's a proof . . .

Let $A$ be a nonempty subset of $X$ such that $A$ has an upper bound. 

Suppose $A$ has no least upper bound.

Our goal is to derive a contradiction.

Let $B$ be the set of upper bounds of $A$.

Since $A$ has an upper bound, $B$ is nonempty. 

Also, by assumption, $B$ has no least element. 

Choose $a_1 \in A$, and $b_1 \in B$.

For each $a \in A$, let $I_a = (-\infty,a) = \{x \in X\mid x < a\}$.

For each $b \in B$, let $J_b = (b,\infty) = \{x \in X\mid x > b\}$.

Let $S = \{I_a\mid a \in A\} \cup \{J_b\mid b \in B\}$.

Let $x \in [a_1,b_1]$.

If $x \in B$, then since $B$ has no least element, there exists $b \in B$ such that $b < x$, hence $x \in J_b$.

If $x \notin B$, then there exists $a \in A$ such that $a > x$, hence $x \in I_a$.

It follows that $S$ is an open cover of $[a_1,b_1]$.

By hypothesis, $[a_1,b_1]$ is compact, hence some finite subset of $F$ of $S$ also covers $[a_1,b_1]$.

Necessarily $F$ contains at least one of the sets $J_b$, else $b_1$ is not covered.

Since $\{b \mid J_b \in F\}$ is finite and nonempty, it has a least element, $b_0$ say.

But then $b_0$ is not covered by $F$, contradiction.

Therefore $A$ must have a least upper bound.

It follows that $X$ has the least upper bound property.
A: $\mathbb{R}$ shows that the statement is false, as you write it.
You can show $X$ has the least upper bound property, though, not that is has one itself. This means that any $A \subseteq X$ that is bounded above has a least upper bound. 
This is why proofs start with $a \in A$ and $z$ an upperbound for $A$. We then need to find a least upper bound for $A$ (not $X$) in the compact interval $[a,z]$ inside which it must lie by definition ($\ge a$ as $a \in  A$ (it's an upper bound) and $\le z$ because it is the least upper bound, and $z$ is one of those upper bounds).
