TITLE: Is convex set in an ordered set necessarily interval or ray? Munkres 16. 7
QUESTION [9 upvotes]: The is Problem 7 in Section 16 (page 92) of Munkres' Topology. The problem reads as follows.

Let $X$ be an ordered set. If $Y$ is a proper subset of $X$ that is
  convex in $X$, does it follow that $Y$ is an interval or a ray in $X$?

The answer is "No." This problem has been discussed here in MSE at least twice, see here and here.
This post is not a duplicate: I think I came up with another counter-example, and I would like to check whether it is correct or not. (I would be proud if it turned out to be right!) Here it goes:

Let $X=\mathbb{R}\times\mathbb{R}$ be equipped with the dictionary order. Consider $Y=\{x\times y: -1\leq x\leq 1\}$. Then $Y$ is convex in $X$, but it
  is not an interval or a ray, as we now prove. If it were an interval,
  it would be of the form $(a\times b, c\times d), (a\times b, c\times
 d], [a\times b, c\times d)$ or $[a\times b, c\times d]$ and none of
  these can be $Y$. Similarly, none of the rays $(a\times b, \infty),
 [a\times b, \infty),  (\infty, a\times b), (\infty,  a\times b]$ can
  possibly equal to $Y$. 

On the one hand, I think this counter-example works; on the other hand, this means in light of Brian's comment, $\mathbb{R}\times\mathbb{R}$ is not Dedekind complete (at least under dictionary order), which is I guess strange. Can someone check whether my counter-example is valid or not? 
Thanks for your time.

REPLY [7 votes]: Your counterexample works, and it shows that $\mathbb{R}\times\mathbb{R}$ is not Dedekind-complete under the dictionary order. Indeed the set $Y$ is bounded - above and below - but it has no least upper bound. $(x,y)$ is an upper bound of $Y$ if and only if $x > 1$, and then $(x,y-1)$ is an upper bound too.