Linear continuum of $I\times I $ under subspace topology of $\mathbb{R}^2$ with dictionary topology on it, where $I=[a,b]$.

Question

Let $I\times I$ be subspace of space $\mathbb{R}^2$ with dictionary order, where $I=[a,b]$. What can you say about the linear continuum of $I\times I$ with the subspace topology.

{ I've proved that $I\times I$ is not connected under the subspace topology(proof is given below check if something is wrong with my proof). But I'm not able to prove that $I\times I$ is not linear continuum (without using the fact that it's not connected). } $\textbf{Proof of $I\times I$ is not connected under subspace topology of $\mathbb{R}^2$ with dictionary order:}$

since $\{x\}\times I$ is open in $I\times I$ for each x in $I$. We can find a separation $$(\{a\}\times[a,b],\bigcup\limits_{y\in(a,b]}\{y\}\times I) $$ clearly $(\{a\}\times[a,b])\bigcap(\bigcup\limits_{y\in(a,b]}\{y\}\times I)=\phi$ and $(\{a\}\times[a,b])\bigcup(\bigcup\limits_{y\in(a,b]}\{y\}\times I)=I\times I$

hence $I\times I$ is not connected.

{$\textbf{Linear Continuum}$(source: Topology 2nd ed. by J. R. Munkres page no. 151): A simply ordered set $L$ having more than one element is called a $\textbf{linear continuum}$ if the following hold:

1. $L$ has the least upper bound property.
2. If $x<y$, there exists $z$ such that $x<z<y.$}

Answer 1

If $\mathbb{R}^2$ has the dictionary order, $I \times I$ as a subset of that ordered plane is not connected (thought not for the reason you write), because each stalk $\{x\} \times I$ for $x \in I$ is closed-and-open. So it's not what Munkres calls a linear continuum, because he shows that those are connected. $\mathbb{R}^2$ also is not a linear continuum in its own dictionary order, as $\{0\} \times \mathbb{R}$ has no least upper bound (but is upper bounded).

OTOH, if we give the set $I \times I$ the order topology wrt the restricted version of the dictionary order, then it is a linear continuum (proved several times on this site, search...). It's a classic example of where restricting the order and taking the topology and restricting the topology from an order have different effects. ($I \times I$ is not order convex, that's part of it.) It's classical that $I \times I$ in the inherited order is a continuum. But its subspace topology is not induced by that order, hence the disconnectedness.