It turns out that the familiar result that surjective monotones $\mathbb R\to\mathbb R$ are continuous extends to general LOTS (linearly ordered topological spaces):
Theorem. If $X$, $Y$ are LOTS with $Y$ being dense and complete, then any surjective monotone $X\to Y$ is continuous.
(The same proof from $\mathbb{R\to R}$ case sails through.)
The necessity of surjectivity is clear by considering the function $f\colon \mathbb {R\to R}$ given by $f(x) := x$ for $x < 0$ and $f(x) := x +1$ for $x\ge 1$.
However, I am having trouble finding two examples that show the necessity of denseness and, respectively completeness of the codomain space. Any leads?
dense-in-itself suffices:
Let $(X, \le), (Y, \le)$ be linearly ordered sets equipped with the order topology, $Y$ dense-in-itself (i.e. for each $y, z \in Y, y < z$, it is $(y, z) \neq \emptyset$.
Let $f: X \rightarrow Y $ be a monotone increasing, surjective map. Then f is continuous.
PROOF. Let $x \in X$, $f(x) \in V$ open in $Y$. Case 1: $f(x)$ is neither minimum nor maximum of $Y$. Then there exist $u, v \in Y$ such that $u < f(x) < v$ and $(u, v) \subset V$. Since $Y$ is dense-in-itself, we may assume that $[u, v] \subset V$. Since $f$ is surjective, pick $a, b \in X$ such that $f(a) = u, f(b) = v$. Hence $a < x < b$ and $f( (a,b) ) \subset [u. v] \subset V$.
Other cases: similar.
It doesn't hold in general, if $Y$ is not dense-in-itself:
$f: \mathbb R \rightarrow \{0, 1\}, \space f(x) = \begin{cases} 0, & \text{if } x < 0 \\ 1, & \text{if } 0 \le x \end{cases} $