I read that the topological space $X=[0,1)\times[0,1]$ with the dictionary order and order topology is not a linear continuum, as it does not satisfy the least upper bound property. (The definition of a linear continuum being a dense linear order with the least upper bound property.)
However, I can't find a nonempty bounded set with no supremum which leads to this violation. My thinking is if $A$ is any subset, and $\pi_1(A)$ is the projection onto the first coordinate, then $b=\sup(\pi_1(A))$ must exist, since if $\pi_1(A)$ is not bounded, then $A$ is not bounded above in $X$. Then the least upper bound of $A$ is $\sup(\pi_1(A))\times \sup(\pi_2(A\cap (b\times [0,1])))$. I feel the only difficulty occurs when $\sup(\pi_1(A))=1$, but then $A$ would not be bounded in the first place, so the situation doesn't apply. Is my thinking wrong, or is $X$ actually a linear continuum?