Assume that the least upper bound property does not hold. This means that there will exist some subset $S$ of $X$ such that $S$ has an upper bound, but not a least upper bound. Any ordered set that contains an upper bounds will have a least upper bound: the upper bound that it contains. Thus the subset $S$ that is a counterexample to the LUB property must not contain an upper bound. Now consider the set $S'$ which is the set of all $s$ such that $s \ \leq \ x$, for at least one $x \ \in \ S$. It is easy to see that this set is open in the order topology on $X$. It also follows from the definition of the set, and more directly, the fact that $S \ \subset \ S'$, that the set of upper bounds of $S$ is equal to the set of upper bounds of $S'$.

Consider the set of upper bounds on $S'$, which we will call $L$. Recall this is also the set of upper bounds of $S$. Since, by assumption, this set does not have a least element, it must not contain its infimum, and $L$ is thus open in the order topology. We assert that $S'$ and $L$ form a separation of $X$. These two sets are disjoint. If they weren't then there would exist some $\ell$ of $L$ such that $\ell \ \leq \ x$, for some $x \ \in \ S$. In the case of equality, we contradict the definition of $S$. Otherwise, we contradict the fact that $\ell$ is an upper bound on $S'$. Now we prove the two sets contain all points of $X$. Assume there is some $z \ \in \ X$ that is not contained in either $S'$ or $L$. It would follow that $z$ is greater than all elements of $S'$, which is a contradiction, as this would be an upper bound of $S'$, and be contained within $L$. Thus, $S' \ \cup \ L \ = \ X$. This contradicts the initial assumption that $X$ is connected. Thus, $X$ must satisfy the LUB property.

Next, we have to prove that for any two $x, \ y \ \in \ X$, there exists some $z$ such that $x \ < \ z \ < \ y$. Assume that this wasn't the case. This means that there would exist $x$ and $y$ with no $z$ such that $x \ < \ z \ < \ y$. Consider the sets:
$$S_y \ = \ \{ s \ | \ s \ < \ y\} \ \ \ \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ \ \ \ \ S_x \ = \ \{s \ | \ s \ > \ x \}$$ Both sets are disjoint and contain all points of $X$, due to the lack of existence of $z$. Both sets are also open in the order topology. Thus, $S_y$ and $S_x$ constitute a separation of $X$, contradicting the initial assumption that $X$ is connected. Thus, $X$ must satisfy the above axiom, and we have shown that $X$ is a linear continuum.