If there were an empty interval $(a,b)$ then $(-\infty,b)\cup(a,+\infty)$ would be a separation, therefore, for any $a<b$ there is $c:a<c<b$ . To show the least upper bound property suppose that $A$ is nonempty and bounded from above. Let $B$ be the set of all upper bounds of $A$ , it is nonempty. All we need to show is that $B$ has the least element. Let $A'=\{x\in X|\exists a\in A:x\le a\}$ . Note that $b$ is an upper bound of $A$ iff for every $a\in A$ : $b\ge a$ iff for every $a\in A'$ : $b\ge a$ iff $b$ is an upper bound of $A'$ . Therefore, $B$ is the set of all upper bounds of $A'$ as well. If $A'$ has the greatest element then it is the least element of $B$ , so, suppose it does not. Then $A'$ is open: if $a\in A'$ then, since there is no largest element in $A'$ , there is $a'\in A'$ such that $a\in(-\infty,a')\subseteq A'$ . Since $X$ is connected, $A'$ is not closed and there is a limit point $b$ of $A'$ in $B$ . It is an upper bound of $A'$ , and suppose there is $b'\in B$ such that $b'<b$ . Then $(b',+\infty)$ is a nonempty open set that contains $b$ but does not intersect $A'$ . This contradicts the assumption that $b$ is a limit point of $A'$ . So $b$ is the least element in $B$ and the least upper bound of $A$ .