Q:

Why does Gauss's law work for a charge off center in a spherical surface?

CASE 1: Consider an enclosed spherical surface with a charge $q$ at its centre. From Gauss' law we can say that the flux through this sphere is $q/\epsilon_0$. 
CASE 2: The charge is inside but off centre. In this case the electric field will not be normal to the surface. So how can we use the formula $q/\epsilon_0$? It turned out that the answer is same for both cases. This is my confusion. 

A:

You can use $q/\epsilon_0$ to calculate flux for both cases because that's what Gauss' law says: Just look at the enclosed charge. It's amazing.
Flux for any closed surface is $\Phi = \oint \vec{E} \cdot d\vec{A}$. There are two ways to calculate this quantity:

The hard way, which means evaluating $\vec{E}$ on every part of the surface, and integrating. This is hard because for your case 2, the electric field strength varies on the surface and it is not normal everywhere. When computed in this manner, it is not obvious at all that your two cases should yield the same value for flux. It really isn't obvious.
The easy way, which is using Gauss' law $\Phi = Q_\text{encl}/\epsilon_0$. To evaluate flux using Gauss' law, you only need to look at how much charge how is physically enclosed. That's what Gauss' law says. That's why Gauss' law gives the same answer in both cases. It makes no mention of the particular geometry of your closed surface and does not assume that the electric field is normal to the surface.

There are proofs of Gauss' law floating around, but I don't think you are asking for that.
As a side note, one often uses Gauss' law to figure out what the electric field is. Case 1 is very useful for figuring out the electric field at some distance away from your charge. Case 2 is not useful since you can't factor out $\vec{E}$ from the integral. Nonetheless, both cases yield the same value for flux.

