Let $X$ be totally ordered by the strict total order $<$ and let it have the order topology.

Suppose $X$ is not a linear continuum. Then either $X$ is not bounded-complete, or the order on $X$ is not a dense total order.

Suppose $X$ is not bounded-complete. Let $A$ be a nonempty subset of $X$ that is bounded above by $b$ but has no least upper bound. Let $U$ be the set of upper bounds of $A$ If $x \in U$ then $x$ is not a least upper bound of $A$ so there is a $z \in X$ such that $z<x$ and $z$ is an upper bound of $A$ Then the set $\{y\in X \mid y>z\}$ is open and contains $x$ Furthermore, all of its elements exceed $z$ so it is a subset of $U$ Thus, $U$ is open. $U$ contains $b$ so it is not empty. Let $x\in X \setminus U$ Then $x$ is not an upper bound of $A$ so there is an $a \in A$ such that $x < a$ The set $\{y \in X \mid y < a\}$ is open, and contains no upper bounds of $A$ so it is a subset of $X \setminus U$ Thus $X \setminus U$ contains a neighborhood of each of its points, and is therefore open. Since $U$ and $X \setminus U$ are open, X is not connected.

Suppose the ordering of $X$ is not dense, so there are $a$ and $b$ in $X$ with $a<b$ so that there is no $c$ in $X$ with $a<c<b$ (there is a gap between $a$ and $b$ . Let $U = \{x \in X \mid x < b\}$ and let $V = \{x \in X \mid x > a\}$ Because there are no elements between $a$ and $b$ $U \cap V = \emptyset$ By transitivity and trichotomy, $U \cup V = X$ $U$ and $V$ are both open. $a \in U$ and $b \in V$ so neither $U$ nor $V$ is empty. Thus, $U$ and $V$ separate $X$ so $X$ is not connected.

Therefore, if $X$ is connected, then $X$ is a linear continuum.

Now suppose that $X$ is disconnected and bounded-complete, and that $U$ and $V$ are (nonempty, open and closed) sets separating $X$ Suppose that $a \in U$ and suppose also that there is an element $b \in V$ such that $a < b$ (if there is none, swap the names of $U$ and $V$ or reverse the ordering). $Z = \{x \in V \mid x > a\}$ is open (it is the intersection of two open sets), and contains $b$ (so it is not empty). $Z$ is bounded below by $a$ so it has a greatest lower bound $z$

If $z \in U$ then, since $U$ is open, there is an interval in $U$ containing $z$ which must, to exclude $b$ be of the form $\{x\mid x<k\}$ or of the form $\{x\mid j<x<k\}$ for some $k$ and perhaps $j$ But then $k$ would be a lower bound of $Z$ contradicting the fact that $z$ is the infimum of $Z$

If $z \in V$ then, since $V$ is open, there is an interval in $V$ containing $z$ which must, to exclude $a$ be of the form $\{x\mid j<x\}$ or of the form $\{x\mid j<x<k\}$ for some $j$ and perhaps $k$ In either case, $a<j<z$ and the set $W=\{x\mid j<x<z\}$ is a subset of $V$ If $x \in W$ then $x \in V$ and $x > a$ so $x \in Z$ contradicting the fact that $z$ is the infimum of $Z$ Thus, there are no elements of $X$ between $j$ and $z$ so the order on $X$ is not dense. This proves that if $X$ is a linear continuum, then $X$ is connected.