Definition: Let $L$ be simply ordered set with more than one element. We say that $L$ is a linear continuum iff. the following two properties holds:
$L$ has the least upper bound property.
If $x,y\in L$ with $x<y$, then $\exists z.\ x<z<y$.
In some sense linear continuums are a generalization of $\mathbb{R}$. And it's easy to see that axioms of $\mathbb{R}$ satisfy the definition of a linear continuum.
But if we take $I^2,$ where $I=[0,1]$ with dictionary order and denote it as $I^2_o$ then I claim that it is also a linear continuum.
The property 2. is easy to check.
Let's show that 1. also holds. If $A\subset I^2_o$ s.t. $A\neq \emptyset$ and $A$ is bounded above by $(x_0,y_0)$ then $\pi_1(A)$ is nonempty and bounded by $x_0$ then it has supremum which we call $l_1$ and using the same for $\pi_2(A)$ we get supremum $l_2$. Then one can show that $A$ is bounded above by $(l_1,l_2)$ and this upper bound is the minimal one.
Sorry that I missed most details. I just want to check is my reasoning ok?
Would be very thankful for any comments!