The proof uses calculus. When we say that pi is equal to the infinite series 4/1 - 4/3 + 4/5 - ..., what we really mean is that the limit of the sum of the series is pi as you increase the number of terms. That means for any number less than pi you can add up a certain number of terms and get a number greater than the original number, but the sum will never be greater than pi.
I don't know how much of calculus you know, so I'll try and explain the proof fully as far as I can. Also note that this isn't really a precise proof that a mathematician would be satisfied with, but it does make sense.
Consider the series 1 + r + r^2 + r^3 + ..., where r is any number. This is called a geometric series, and as long as |r| < 1 so that the terms get smaller and smaller, the infinite sum of it (the limit of its sum as the number of terms increases) is 1 / (1 - r). To prove this, consider a finite part the series, 1 + r + r^2 + ... + r^n. Let's call the sum of this part S, so
S = 1 + r + r^2 + ... + r^n
Now, if you multiply S by r, you can spread the r through S and see that
r S = r (1 + r + r^2 + ... + r^n) = r + r^2 + r^3 + ... + r^(n + 1)
This is just like the original part of the series we had, except that we've lost the first term and added on an additional term. 1 is a term in S but not in r S. r^(n + 1) is a term in r S but not in S. All the other terms are common to both. And we can subtract the two sums, and all the common terms cancel out.
S - r S = 1 - r^(n + 1)
Now we simply rearrange this to find S.
S(1 - r) = 1 - r^(n + 1)
S = (1 - r^(n + 1)) / (1 - r)
That gives us a formula for the sum of 1 + r + r^2 + ... + r^n for any value of n. Now, suppose |r| < 1 and also n is infinite. If you think about a number like 0.5, multiplying by itself halves it to 0.25, multiplying it by itself again halves it to 0.125, and so on. Whenever a number's magnitude is less than 1, multiplying something by that number makes it smaller. So it's fairly easy to see that whenever |r| < 1, the limit of r^n as n gets larger and larger is 0. This is a part where a mathematician would want a more precise proof, but it makes sense.
Using that fact, we can then say that the limit of (1 - r^(n + 1)) / (1 - r) as n approaches infinity would be 1 / (1 - r) since the subtracted r^(n + 1) part would approach zero. And we know that that limit is also the limit of 1 + r + r^2 + ... + r^n as n approaches infinity, i.e. the value of the infinite sum 1 + r + r^2 + ... This proves that
1 / (1 - r) = 1 + r + r^2 + ... when |r| < 1.
This is an example of how you can carry out reasoning about infinite sums. Now, it still seems unclear how we get to pi from here. But we can do it. The key is to substitute the value -x^2 in for r. Then you get the equation
1 / (1 - (-x^2)) = 1 - x^2 + (-x^2)^2 + ... when |-x^2| < 1
1 / (1 + x^2) = 1 - x^2 + x^4 - ... when |x| < 1
Now, this is the part where you really need some calculus. I don't know how much you know about integration. But every function has a related function called an integral. If you plot the integral on a graph, then the value of the original function at every point gives you the gradient (or you might call it slope, if you're American) of the graph of the integral at that point. The original function is known as the derivative.
The integral of 1 / (1 + x^2) turns out to be arctan x (when the units of x are radians). It would probably be too complicated to fully explain this here (but it's not at all hard to prove when you are familiar with calculus). Also, the integral of a sum is the sum of the integral of all the terms, so we can find the integral of 1 - x^2 + x^4 - ... by finding the integral of each term. There is a simple formula for this: the integral of x^n where n is any integer is x^(n + 1) / (n + 1). This means the integral works out as
x - x^3 / 3 + x^5 / 5 - ...
And it stands to reason that if two functions are equal, their integrals should also be equal. So we can now say that
arctan x = x - x^3 / 3 + x^5 / 5 - ...
Now it looks like we're getting somewhere. And tan (pi/4) is 1 (since pi/4 in radians is 45 degrees), so arctan 1 = pi/4, meaning
pi/4 = 1 - 1 / 3 + 1 / 5 - ...
Then if we multiply the equation through by 4 we get
pi = 4 - 4 / 3 + 4 / 5 - ...
There is a problem here: that series for 1 / (1 + x^2) was only valid when |x| < 1, and to get this series for pi we have had to make x equal to 1. The series is actually still valid, but strictly speaking we haven't proved that yet. I don't know any convenient way of proving that from this point. (You can take an alternative approach and find the series for arctan x using Taylor's formula, which will not involve any requirements that |x| < 1 or anything).