Prob. 4, Sec. 24 in Munkres' TOPOLOGY, 2nd ed: How to verify the supremum property? Let $X$ be an ordered set in the order topology. Suppose that  $X$ is connected. 
How to show that $X$ is a linear continuum? 
 A: Suppse $x, y \in X$ such that $x < y$. If $(x,y)$ were empty, then we can write $X$ as
$$X \ = \ (-\infty, y) \cup (x, +\infty),$$
which implies that $X$ is not connected. So there exists $z \in X$ such that $x < z < y$.
Now let $A$ be a non-empty subset of $X$ such that $A$ is bounded above in $X$. Let $U(A)$ be the set of all the upper bounds of $A$ in $X$. Then $U(A)$ is also a non-empty subset of $X$. We need to show that $U(A)$ has a smallest element.
If $A$ has a largest element, then that largest element is also the smallest element of $U(A)$ and hence the supremum of $A$. So we assume that $A$ has no largest element.
Assume also that $U(A)$ has no smallest element. Then if $u \in U(A)$, then we must have $u^\prime < u$ for some element $u^\prime \in U(A)$, so that $u \in (u^\prime, +\infty) \subset U(A)$ and $(u^\prime, +\infty)$, being an open ray, is a subbasis element for the order topology on $X$. Thus it follows that the set $U(A)$ is a non-empty open subset of $X$.
Now if $x \in X - U(A)$, then $x$ is not an upper bound for $A$; so there is some element $a = a(x) \in A$ such that $x < a(x)$. Since $A$ has no largest element, the intersection of $A$ and $U(A)$ is empty. So this particular element $a(x) \not\in U(A)$ and hence $a(x) \in X-U(A)$.
Thus, for every $x \in X-U(A)$, we have $x \in \left( -\infty, a(x) \right) \subset X-U(A)$, and the open ray $\left(-\infty, a(x) \right)$ is a subbasis element for the order topology on $X$.
Thus $X-U(A)$ is a non-empty open subset of $X$ as well.
Therefore, $U(A)$ and $X-U(A)$ is a separation of $X$ and so $X$ is not connected, a contradiction.
