Proof that $\nabla \times E = 0$ using Stoke's theorem One way that Jackson proves that $\nabla \times E = 0$ is the following:
$$ F = q E $$
$$ W = - \int_A^B F \cdot dl = - q \int_A^B E \cdot dl = q \int_A^B \nabla \phi \cdot dl = 
 q \int_A^B d \phi = q(\phi_B - \phi_A) $$
so $\int_A^B E \cdot dl = -(\phi_B - \phi_A)$ therefore $\oint_A^B E \cdot dl = 0$
Then using Stoke's theorem $\int (\nabla \times E) \cdot \hat{n} da = 0$ which implies $\nabla \times E=0$.

*

*Isn't the last line true only if $\nabla \times E \geq 0$?

*How does $\nabla \phi \cdot dl = d\phi$ in general? I can only see how this is true if $dl = dx \hat{x}$ or similarly for $y$ or $z$.

*The initial integration for $W$ is over a path in space (w.r.t $x,y,z$), yet at the end we are integrating w.r.t $\phi$? How does that even make sense?

I was split on whether to post this here or on the Math site. I figured maybe the answer to my first question could only be due to the specifics of $\nabla \times E$, since it seems like Jackson is instantiating $\int f(x) dx = 0 \implies f(x) = 0$, but this isn't guaranteed unless we already know $f(x) \geq 0$.
 A: 
...
Then using Stoke's theorem $\int (\nabla \times E) \cdot \hat{n} da = 0$ which implies $\nabla \times E=0$.



*

*Isn't the last line true only if $\nabla \times E \geq 0$?


No. It isn't. The closed loop path is arbitrary.
Also, you don't need Stokes' theorem. You already have $\vec E = -\vec \nabla \phi$, so:
$$
\vec \nabla \times \vec E = -\vec \nabla \times \vec \nabla \phi
$$
$$
=\hat e^{(i)} \epsilon_{ijk}\nabla_j\nabla_k\phi = 0\;,
$$
by the symmetry of $\epsilon_{ijk}$ (totally antisymmetric in all indices) and $\nabla_j \nabla_k \phi$ (totally symmetric in all indices).



*How does $\nabla \phi \cdot dl = d\phi$ in general? I can only see how this is true if $dl = dx \hat{x}$ or similarly for $y$ or $z$.


Or similarly, in general:
$$
d\phi = \frac{\partial\phi}{\partial x}dx
+\frac{\partial\phi}{\partial y}dy
+\frac{\partial\phi}{\partial z}dz
$$
$$
=\vec \nabla\phi\cdot d\vec{\ell}
$$



*The initial integration for $W$ is over a path in space (w.r.t $x,y,z$), yet at the end we are integrating w.r.t $\phi$? How does that even make sense?


Because you changed integration variables from $\vec{\ell}$ to $\phi$.
This, again, is basically the why we introduce an integral in terms of the anti-derivative. It the analog of the 1-d example:
$$
f(x) = \frac{dF}{dx}
$$
and
$$
\int_{x_1}^{x_2} dx f(x) = \int_{F(x_1)}^{F(x_2)} dF = F_2 - F_1\;.
$$

All these question seem to be questions about basic or intermediate vector calculus. I suggest you find an intermediate level undergraduate textbook on vector calculus and refresh your memory as to these fundamental concepts.
