As long as connectedness implies linear continuum, then I am induced to think that all intervals of a linear continuum inherit the linear continuum property, and as long as it is an iff statement, they are connected. Can I generalize the proof of $\mathbb{R}$ for a general connected totally ordered space?

Proof:
By contradiction, assume that $J:=[x'',x'] \subseteq X$ is an interval and it is not connected. Let $A,B$ be two nonempty and disjoint closed subsets of $J$ such that $J= A \cup B$. Pick $x \in A$ and $y \in B$ and assume, without loss of generality, that $x \prec y$. As $J$ is an interval, we have $[x,y] \subset J.$ Denote now $U:= A \cap [x,y)$ and $z:= \sup U$. As $U \subset [x,y]$ and $[x,y]$ is closed in $X$ and contained in $J$, it follows that $\overline{U} \subset J$ and so $\overline{U}$ is a closed set in $J$. As $U \subset A$ and $A$ is a closed subset of $J$, it follows that $\overline{U} \subset A$. We can now write that $z \in \overline{U}$ and so $z \in A$. As $z \in [x,y] \cap A$ and $y \in B$, from $A \cap B = \emptyset$ we can write $y \succ z$. Moreover, as $z$ is the supremum of the elements of $A$ that are less than $y$, it follows that $(z,y] \subset B$ and so $z \in \overline{B}$. But now, as $z \in J$ and $B$ is a closed subset of $J$, we have that $z \in \overline{B} \cap J = B$, reaching a contradiction.