Why does Gauss's law apply to any shape of a closed surface? What seems to incredibly bother me is why Gauss's law applies to any shape of a closed surface. Moreover, the fact that the electric flux is proportional to the enclosed charge is by many sources simply proven by using a point particle that is enclosed by a sphere. Hence the electric flux is proportional to the enclosed charge for any closed surface, which to me isn't obvious by just using a specific proof that includes merely a sphere.
Furthermore I have watched various videos on youtube, including a lecture given by Walter lewin, and visited numerous websites, however, all sources didn't resolve my confusion.
Lastly, why does Gauss's law also work for any collection of randomly distributed charges? Many sources state the any collection of charges can be thought of a collection of separate point charges, and since the electric fields of point charges should be added vectorially, they can also be be considered to be one total charge that generates one net electric field. Does this imply that the closed surface integral, used in Gauss's law, can be divided into separate closed surface integrals? Like so:
$$ \int\sum\ \vec E\bullet d\vec A =  \int\ \vec E1\bullet d\vec A +\int\ \vec E2\bullet d\vec A
+ \ldots +\int\ \vec Ei\bullet d\vec A$$
Where every seperate line integral includes the electric field of a single point charge.
Now my mathematical toolkit is relatively limited, thus, please do not utilize complicated mathematical equations that originate from divergence, differential equations, etc.
Thank you in advance.
 A: If you are convinced by the sphere argument, then consider for any irregular surface a spherical surface within it that contains entirely your charge. Then within a "sector" of certain solid angle, the field lines that pass through the intersection of this "sector" and your irregular surface are exactly those that pass through the projection of the irregular surface onto the sphere, and so the total number must be exactly the same.
If you want a more quantitative (read: rigorous) solution, I will have to direct you to the divergence theorem, there is just no way around it. To be frank, if you don't want to learn vector calculus, you will probably just have to accept some results for what they are.
Also, yes, the closed surface integral of the sum can be expressed as the sum of the integrals.
A: The integrals are not “line integrals”; they are an expression of flux (field x area).  If you divide the enclosed volume into many very small solid angles, the dot product means you can use the component of the area at the outer end of the solid angle (represented by a vector pointing out) which is parallel to the field (and the radius).  That area increases with the square of the radius and the field decreases with the square of the radius.  That means that the flux depends only on the charge and the size of the solid angle; and not the radius or the slant of the actual surface.  The integral involves summing the solid angles (the area of a sphere of any radius divided by the radius).
