Connected subsets of lines For me every connected set has at least two elements.
Connected subsets of the real line have non-empty interiors. I am curious if the same is true for connected subsets of any connected linearly ordered topological space (the topology comes from the order).

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.
