Gaussian curvature of a surface which is equal to the $(x,y)$-plane outside a ball of radius $10$ The problem I'm working on is as follows

Let $M \subseteq \mathbb{R}^3$ be a non-compact orientable surface without boundary which coincides with the $(x, y)$-plane outside of the ball of radius $10$ centered at the origin. Prove that if the Gaussian curvature $K$ of $M$ is everywhere non-negative, then $K$ is everywhere $0$.

A few days ago, I posted here an idea I had for this problem, hoping someone could help me flesh out the missing step(s). The idea I had was to look at the part of the surface contained in the ball of radius $10$, call it $R$, a surface w/ boundary, and apply Gauss-Bonnet to it to establish that $\iint_R K \mathrm{d} M \leq 0$. This would imply that $K$ is identically $0$. I later realized that this method won't work, since we can't guarantee that $R$ is compact, a necessary assumption to use GB. For example, if $M$ was just the $(x, y)$-plane minus the origin, it would fit the hypotheses of the problem, but $R$ would be a punctured disc, which is not compact.
My second idea had been to look at the point in the surface where the $z$-coordinate was maximized and see if I could come to some conclusion about the curvature there, but this won't work for similar reasons, since the $z$-coordinate could be unbounded. I have in mind something where there's a "singularity" about the $z$-axis.
So I'm out of ideas for how to solve this problem. I don't really know what's left to try. I would love hints, but I'm studying for an exam in two days, so I'd also appreciate worked-out solutions to this problem.
Thanks!
 A: I think that even if the surface is assumed to be connected, the statement is false.
Let $\phi:(-1,1)\to\mathbb R$ be a smooth function satisfying $\phi(x)=0$ for $|x|\leq\frac 12$, $\phi''(x)\geq 0$, $\phi''(x)>0$ for $|x|>\frac 12$ For example $\phi(x)=\exp(\frac{-1}{|x|-\frac 12})$ for $|x|\geq\frac 12$ and $0$ otherwise.
On $R=(0,1)\times (-1,1)$ define $f(x,y)=\phi(x)+\phi(y)$. The graph of $f$ looks like this:

The curvature is given by
$$K(x,y)=\frac{f_{xx}f_{yy}-f_{xy}^2}{(1+f_x^2+f_y^2)^2}
=\frac{\phi''(x)\phi''(y)}{(1+\phi'(x)^2+\phi'(y)^2)^2}\geq 0$$
and we have $K(x,y)>0$ for $|x|,|y|>\frac 12$. Furthermore $f=0$ on $(0,\frac 12)\times (-\frac 12\times\frac 12)$.
Now we let $\Omega\subset\mathbb R^2$ be the union of $\mathbb R^2$ with the closed unit disk of radius $10$ removed, the stripe $(-\infty,\frac 12)\times (-\frac 12,\frac 12)$ and $R$. On $\Omega$ we define $F$ to be $f$ on $R$ and $0$ otherwise. Then $F$ is smooth and the graph of $F$ is the desired counterexample.
$\textbf{In summary:}$ If the closed unit disk of radius $10$ is a sea in $\mathbb R^2$, then we put our little boat from the picture in the middle of the sea and add a flat bridge of width $1$ to the land.
A: As stated, it's false, but for stupid reasons. Consider the union of the xy-plane with the radius-$10$ disk deleted, and the upper hemisphere of a sphere of radius $10$ at the origin. That's a surface, and at no point of that surface is there negative curvature. (Along the equator, the surface is not smooth, so the curvature is undefined, but that doesn't make it negative.) But at the north pole, the curvature is clearly strictly positive.
This makes me think that the examiners were not being very careful in writing the question. I expect that the "noncompactness" was meant to apply to the fact of the surface matching the $xy$-plane, and extending to infinity, while inside the ball of radius $10$, they probably did mean to assume compactness, allowing the Gauss-Bonnet proof.
