Problem: $X$ is an ordered set with order topology. Is it true that (1) $A\subseteq X$ is connected $\implies$ $A$ is convex (2) $A\subseteq X$ is connected $\implies$ $A$ is an interval ? (Here interval can be open, closed, half-open half-closed, and boundary can be $(-\infty$ or $+\infty)$, or has one element since $\{a\}=\{x\ :\ a\le x\le a \}$).
Motivation: In the order topology of $\mathbb{R}$ (which is the same as usual topology on $\mathbb{R}$), it is easy to see that $A\subseteq \mathbb{R}$ is connected $\iff$ $A$ is convex $\iff$ $A$ is an interval. I want to know if this holds for general order topology. Now I have worked out four arrows:
(1) Convex does not imply connectedness. counterexample: the order topology on $\mathbb{Z}_{\ge 0}$ with subset $\{2,3,4\}$. $\{2,3,4\}=\left(\{2,3,4\}\cap\{x\in\mathbb{Z}_{\ge 0}\ :\ 1< x<3\} \right) \cup\left( \{2,3,4\}\cap\{x\in\mathbb{Z}_{\ge 0}\ :\ x>2\}\right) $ and is hence not connected.
(2) Convex does not imply being an interval, counterexample: the order topology on $\mathbb{Q}$ and the subset $\mathbb{Q}\cap (\alpha, \beta)$, where $\alpha$, $\beta\in\mathbb{R}-\mathbb{Q}$.
(3) Being an interval implies convex: obvious.
(4) Being interval does not imply connectedness: the counterexample in (1) again works.
I did not work out the two arrows in my question.
With $(X, \leq)$ in the order topology, $A \subseteq X$ connected does imply convexity. For if $A$ is not convex, there are $a < b < c$ of $X$ s.t. $a, c \in A$ but $b \not\in A$. So $A \cap (-\infty, b), A \cap (b, +\infty)$ form a disconnection of $A$.
However, $A$ connected does not imply $A$ is necessarily an interval. Indeed for a counterexample consider $X = (\mathbb R^2, \leq_{\text{dict}})$ in the dictionary order and $A = \{0\} \times \mathbb R$. The latter is connected but you can't write it as an interval since "it has no endpoints in $(\mathbb R^2, \leq_{\text{dict}})$" so to speak.