Why is the y-component of electric field of a uniformly-charged disk near its center the same as that of infinite sheet of charge? This is an excerpt from Edward M. Purcell's Electricity & Magnetism:

As $y$ approaches zero from the positive side, $E_y$ approaches $\dfrac{\sigma}{2\epsilon_0}$. On in the negative side of the disk, which we will call the back, $\mathbf{E}$ points in the other direction & its $y$ component $E_y$ is $-\dfrac{\sigma}{2\epsilon_0}$. This is the same as the field of infinite charge-sheet of charge density $\sigma$. It ought to be, for at points close to the center of the disk, the presence or absence of charge out beyond the rim can't make much difference. In other words, any sheet looks infinite if viewed from close up Indeed, $E_y$ has the value $\dfrac{\sigma}{2\epsilon_0}$ not only at the center but all over the disk.

My questions are:
1)Why is the field near the center the same as that of infinite charge-sheet? What is about rim?? I am not getting what logic he is imparting. Can anyone help me explain Mr. Purcell's explanation?
2) What is the proof that all $y$ components of electric field is $\dfrac{\sigma}{2\epsilon_0}$? Why should it be so?
 A: You've got some cgs Gaussian and some mks SI going on in the same problem again in your excerpt, but I'll assume you've figured out how to deal with that since it didn't affect your questions.
Imagine an infinite sheet. The field at a distance y from the sheet equals
$$ \vec E = E_y \hat y =\int_{r=0}^{r=\infty}\frac{\sigma 2\pi r dr}{4\pi\epsilon_0(y^2+r^2)}\frac{y}{\sqrt{y^2+r^2}},$$
where $r, y, \phi$ is a cylindrical coordinate system, we are integrating the plane $y=0$ and $dr$ is covering an annulus in the plane $y=0.$ and we are finding the field on the axis.
Because the integrand is either positive or negative (depending on $\sigma$) but not both and because it integrates to a finite number then you can get within 1% (or 0.00001%) by only integrating over a finite disk.
That's just what an improper integral means, taking the limit of integrals with a finite bound, in this case, the limit of the integrals over disks.
You can express the radius of that disk in terms of y, express it as an angle. So for any level of accuracy there is an angle and closer than making that angle means you have a big enough disk around you for your distance to be within your accuracy for that distance you are close to the infinite sheet example.
So for a finite disk, you can pick a point on the disk, any point that isn't the edge and then zoom in to a point really really close to the finite disk where the angle to be within 1% contains regions all inside the disk, since there is more charge than that you are actually a little closer than 1%. Then you can zoom in more (move closer), to a point really really close to the finite disk where the angle to be within 0.0001% contains regions all inside the disk, since there is more charge than that you are actually a little closer than 0.0001%. And then you can move even closer.
So the field very near the disk is the same as that for an infinite sheet.
When the rim is mentioned it's just that there is no charge beyond the rim. But we don't need to look beyond the rim because you can consider a smaller disk centered around the projection of a point and that disk generates a field really close to the infinite sheet field for a point above its center if the point above the center is close enough, and the other charge within that big disk just makes you a little bit closer in overall magnitude.
