Not really homework but it's a fairly boring question so I'll put it here to avoid annoying people too much.

How would you go about proving that the total solid angle from a closed surface at a point is zero if the surface doesn't enclose the point.

This is obviously needed for Gauss's theorem in EM, but all the books I've got just assert it's true. It seems obviously true from common sense but I'm not sure if it's mathematically trivial.
Thanks

I don't think that this is required for Gauss' theorem. But, anyway...

This is no doubt an inelegant proof, but I think it does qualify as a proof:
Break the closed surface into two open surfaces, separated by the curve at which the dot product between the surface vector and position vector vanishes. integrate the solid angle of each of these surfaces and show that it is the same. Argue that the sense of the surfaces is opposite, and so they cancel.

That'll be hard to make rigorous. It's tough to show that the curve along which you're splitting will have all the necessary good properties.

I'll be a little more verbose.

Your "curve" need not even be a curve. The surface coul d have little plates in it where the surface vector and position vector are normal. God only knows what other kinds of singular behaviour it can have, and what that can mean for the bits on either side.

You'll need to use the fact that a closed curve in a closed surface in R^3 splits it into two bits, which is not trivial; that's only the beginning.

These things are naturally handled by Stokes's theorem. Anything you can produce here by doing your argument will basically be reproving a special case of Stokes's theorem, with all the attendant analytic pain.