Is the preimage of a closed interval under a nondecreasing continuous bivariate function a convex set?

Question

Let $f : \mathbb{R}^2 \to \mathbb{R}$ be continuous and nondecreasing in the sense that $x_1 \leq y_1$ and $x_2 \leq y_2$ implies $f(x_1, x_2) \leq f(y_1, y_2)$. If $I$ is a closed interval, is its preimage $f^{-1}[I]$ necessarily a convex set?

I get the feeling this is false, but I'm not entirely sure how to go about constructing a counterexample. Clearly, if it holds, then $f^{-1}[a]$ must be a (possibly degenerate) line segment for any real $a$. Certainly, it holds whenever $f(x, y) = g(x + y)$ where $g : \mathbb{R} \to \mathbb{R}$ is nondecreasing and continuous.

Answer 1

No, see for instance $f(x,y)=x^3+y^3$. $\{(0,1),(1,0)\}\subseteq f^{-1}[\frac12,\frac32]$, but $(\frac12,\frac12)\notin f^{-1}[\frac12,\frac32]$.