By an ordered structure I mean a (first order) structure $\mathcal{M}=(M,<,\ldots)$ that is totally and densely ordered by $<$. An ordered structure $\mathcal{M}$ is o-minimal if every definable subset $A\subset M$ is finite union of points and intervals.
We consider the order topology in the ordered structure $\mathcal{M}$ and say that $A\subseteq M$ is definably connected if there are no definable open subsets $U_1$ and $U_2$ such that $U_1 \cap U_2 \cap A= \emptyset$ and both $A\cap U_1$ and $A\cap U_2$ are nonempty.
The above definitions are taken from Sergei Starchenko's online notes on o-minimality. Firstly I'd like to comment that the definition for definably connected seems to be missing $A\subseteq U_1 \cup U_2$. Secondly I would like to proof the following:
Every interval in an o-minimal structure is definably connected.
Now in these notes an interval in an ordered structure $\mathcal{M}$ is defined as an interval with endpoints in $M\cup \{+\infty,-\infty\}$. The statement can then easily be proved. If we however considered an interval defined more generally as a set $I\subseteq M$ such that when $x,y\in I$ and $x<z<y$ then $z\in I$, is it still true?
The same proof that works for intervals in the usual sense also works for intervals in your sense. Assume $A \cap U_1 \neq A,\emptyset$. Then there are points $a_1 \in A \cap U_1, a_2 \in A \setminus U_1$. By o-minimality, $U_1$ is a finite union of open intervals, so there is a point $a^\star$ between $a_1$ and $a_2$ which is an (open) endpoint of $U_1$. By your notion of interval, $a^\star \in A$. In order to cover $a^\star$ with a definable open set, by o-minimality, we have to include an open interval containing $a^\star$. It is clear that any such interval must intersect $U_1 \cap A$.
Also, for your further edification, note that, by an easy compactness argument, intervals in your sense are just intervals with endpoints in an elementary extension (or, rather, the intersection of such an interval with the original model).