I am looking to show that $\int_S(\nabla\times$F$).d$A $=0$ for a closed surface S. This is straightforward to do by the Divergence Theorem, but I am being asked to do it by Stokes' Theorem. So, from
$\int_S(\nabla\times$F$).d$A $=\int_C$ F$.d$x,
it makes sense for the RHS to be zero, since (I think) a closed surface S does not have a boundary curve (or the boundary curve is just some singular point?) but I have no idea how to make this rigorous. Do I have to parameterise $C$ in some way? It does not feel complete to me to just say that this is zero because the boundary 'doesn't exist'.
You can use the divergence theorem if your surface $S$ can be viewed as boundary surface $S=\partial D$ of a compact domain $D$, and ${\rm curl}({\bf F})$ is $C^1$ in a neighborhood $\Omega\supset D$. In such a case $$\int_S{\rm curl}({\bf F})\cdot d\vec\omega=\int_D{\rm div}\bigl({\rm curl}({\bf F})\bigr)\>dV=0\ .$$
Now, if $S$ is closed (and orientable) you can triangulate it such that it is nicely covered by surface triangles meeting each other along their edges. The triangles are then oriented coherently, such that along each meeting edge the directions coming from the meeting triangles are opposite. For each triangle $T_i$ we have Stokes theorem, such that we can say that $$\int_S{\rm curl}({\bf F})\cdot d\vec\omega=\sum_i\int_{T_i}{\rm curl}({\bf F})\cdot d\vec\omega=\sum_i\int_{\partial T_i}{\bf F}\cdot d{\bf x}=0\ ,$$ since all occurring edges are counted twice in opposite directions.