Consider two simple closed geodesic curves $\gamma_1$, $\gamma_2$ on a compact hyperbolic surface. Show that $\gamma_1\sqcup\gamma_2$ cannot be the boundary of an (embedded) cylinder.
An embedded cylinder is just a subsurface homeomorphic to $S^1\times[0,1]$. I think this has something to do with the fact that the cylinder has zero curvature, whilst the surface admits a metric with curvature $-1$, but I'm not sure.
Assume $R_1$ is an embedded cylinder in the given compact hyperbolic surface, such that $R_1$ is homeomorphic to $S^1 \times [0,1]$ and the two boundary curves $\gamma_1,\gamma_2$ are geodesics in the given surface.
As you say in the comments, by applying the Gauss-Bonnet theorem it follows that the area of $R_1$ equals $-2\pi \chi(R_1)$.
Since $\chi$ is a homotopy invariant, it follows that $\chi(R_1)=\chi(S^1 \times [0,1]) = \chi(S^1)=0$ (or you can verify more directly that $\chi(S^1 \times [0,1])=0$, if you like). Thus, the area of $R_1$ equals zero.
However, the area of $R_1$ is equal to the area of its interior $R_1 - (\gamma_1 \cup \gamma_2)$ which is a nonempty open subset of the given surface, and the area of any nonempty open subset is positive.
But zero is not positive, and so we have a contradiction.
Regarding your comment that you did not know what kind of contradiction to aim for, sometimes what happens is that you do not have to know because a contradiction will ensue all on its own as long as you follow your nose through the proof. In this case, based on your experience of Euler characteristic, at some point your nose should have pointed towards the equation $\chi(R_1)=0$, and you can then just keep following it.