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.
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.