Improper use of Stokes and Divergence Theorem. Find the problem Could someone point out what is wrong with this equality? Assume that $\mathbf{F}$ is continuous (and hence, its partial derivatives). 
$$\begin{align}
\oint \mathbf{F}\cdot d\mathbf{s} & =^\text{by Stokes} \iint_S \nabla \times \mathbf{F} \cdot d\mathbf{S} \\
&=^\text{by Div} \iiint_V \nabla\cdot( \nabla \times \mathbf{F} ) \, dV \\
&=\iiint_V 0 \,dV \\
&=0\\
&\implies \oint \mathbf{F}\cdot d\mathbf{s}= 0 \; \forall \mathbf{F}
\end{align}$$
Since we assumed $\mathbf{F}$ and its partials are all continuous. But obviously this is wrong if $\mathbf{F}$ is non-conservative. But everything seems to agree. What went wrong?
EDIT. For a refinement of the problem. Let me specifically state that $S$ is a closed surface with a boundary curve that is also closed. So $V$ here is the volume of that surface and since $S$ is closed it has a volume
 A: Actually nothing is wrong with that.  You start with a vector field integrated over a closed curve.  Your first equality which does use Stokes's Theorem goes to an integral over a surface S for which your original curve must be the boundary. Your next equality uses the divergence theorem and goes to an integral over a volume for which your surface S must be the boundary implying S is a closed surface.  Since your assumptions indicate that S is a closed surface S doesn't have a boundary- or rather, the boundary of S is the empty set.  So the integral you started with is over the empty set----> hence it's zero.
A: This is one of my favorite phenomena in multivariable calculus.  I remember noticing this when I was first learning the subject, and spent many hours wondering how this could be.
The explanation for this phenomenon lies in the following geometric principle:** 

The boundary of a boundary is empty.

This geometric fact is in some sense "dual" to the fact that $\text{div}(\text{curl}\,\mathbf{F}) = 0$ for all $\mathbf{F}$.
In particular, if you have a volume $V$ that bounds a surface $S$, then the surface $S$ cannot have a boundary curve.  Said another way, the boundary curve $C = \partial S$ is the empty set, so integrating anything over it is zero.
Example: In the comments, you consider a solid hemisphere $V$.  The boundary of $V$ will then be the surface of the hemisphere and also the disc base.  This closed surface (consisting of both the hemisphere surface and the disc base) does not have a boundary curve.
On the other hand, the surface which is just the hemisphere (without the base) does have a boundary curve: namely, the circle.  However, this surface cannot be said to enclose any volume.

Note 1: Typically when one talks about a "closed" surface, one specifically means a surface which does not have a boundary curve.  This is an unfortunate piece of terminology since the term "closed" can also refer to being a closed subset of $\mathbb{R}^3$, and these two definitions are not equivalent.
For instance, the hemisphere together with its boundary curve (but not including the disc base) is a closed as a subset of $\mathbb{R}^3$, but is generally not called a "closed surface."  However, the hemisphere together with the disc base is a closed surface (and is also closed as a subset of $\mathbb{R}^3$).
** Note 2: This principle is somewhat vague as stated.  In order to make it precise, one needs to rigorously define the notion of "boundary."  This can be done in a couple of ways; some definitions will satisfy this principle, while others won't.  For now, let's not get into these details.
