 Solution

If two simple closed geodesics were disjoint on a sphere-like surface, they would bound an annulus. The Gauss-Bonnet theorem applied to it gives ${\int }_{A}KdA+{\int }_{\partial A}{k}_{g}ds=2\pi \chi (A)=0$. Both boundary geodesic curvatures vanish, but positive Gaussian curvature makes the area integral strictly positive, a contradiction. Hence the two geodesics must intersect.

The assertion fails for nonzero constant signed geodesic curvature. On the unit round sphere a latitude of colatitude $\theta$, oriented in increasing longitude, has $k_g=\cot \theta$. The latitude of colatitude $\pi -\theta$ has the opposite value; reversing its orientation gives the same value as the first. Choosing $0<\theta <\pi /2$ produces two disjoint smooth simple closed curves with the same positive nonzero curvature. Reversing both handles negative values, and $\theta$ can be chosen for any specified nonzero $\lambda$. Thus constant nonzero geodesic curvature does not force intersection.