Yes. In fact, the space need not be connected, as all connected subsets of a linearly ordered topological space are intervals, and in a linearly ordered space connected intervals have nonempty interior. Let $X$ be such a space. To prove the first statement, assume $S$ is connected but not an interval, so we have some $a,b\in S$ and $x\in X$ such that $a<x<b$ but $x\notin S$. Then the sets $(-\infty, x)\cap S$ and $(x,\infty)\cap S$ are open in $S$, nonempty, and their union covers $S$, contradicting the fact that $S$ is connected. To prove the second, let $I$ be a connected interval in $X$ with at least two points $a,b\in I$, chosen so that $a<b$. Then the open set $(a,b)$ is contained in $I$, and the set $(a,b)$ cannot be empty, as otherwise we would have that $$I = \big((-\infty,b)\cap I\big)\cup\big((a,\infty)\cap I\big)$$ is disconnected as these are disjoint nonempty relatively open sets. Thus $I$ has nonempty interior.