I'm working on the following question:
Is every interval in the Dictionary Ordering on $\mathbb{R}^2$ equipped with the order topology connected?
My answer is no: Let $[a,b]$ be an interval where $a = (1,0) \in \mathbb{R}^2$ and $b=(3,0) \in \mathbb{R}^2$. Then sets $$A = \{1\} \times \mathbb{R}_+ \, \bigcup \, (1,2] \times \mathbb{R} \\ B = \{3\} \times \mathbb{R}_- \, \bigcup \, (2,3) \times \mathbb{R}$$ form a separation of $[a,b]$ (both sets are open in the this topology).
However, I'm worried, because dictionary Ordering on $\mathbb{R}^2$ is a linear continuum - and all such sets with the order topology are connected.
What's the issue here?
You wrote yourself in the comments where is that the problem lies: $\mathbb{R}^2$ with the dictionary order is not a linear continuuum. For instance $\left\{(0,x)\,\middle|\,x\in\mathbb{R}\right\}$ has an upper bound ($(1,0)$, for instance), but it has no supremum.