This is a *proof* that a compact field is finite, so you might want to read that first. To help you compare this proof to the NSA proof that a compact field is finite (which uses non-standard analysis), I have endeavored to use similar terms in both.

Let F be a compact field, and suppose to the contrary that F is infinite. Take an infinite sequence of different elements of F. Applying compactness, we see that that sequence must have a subsequence y_{n} which converges to a limit. Denote x = lim_{n→∞}y_{n}. Define z_{n} = y_{n}-x. Then (subtraction is continuous in a topological field) lim_{n→∞}z_{n} = 0.

Now define w_{n}=z_{n}^{-1} (the sequence of multiplicative inverses of the z's). F is compact, so there exists a convergent subsequence of w_{n}: say v = lim_{k→∞}w_{nk} exists, for some increasing sequence of indices n_{k}. By continuity (recall that z_{nk}→0),

0*v = lim_{k→∞} z_{nk}*w_{nk} =
lim_{k→∞} 1 = 1

so 0*v=1. But

this cannot be! (For instance, we deduce that 1=0*v=(0+0)*v=2*(0*v)=2, which is

false).

So our assumption that F is infinite cannot hold; a compact field F must be finite.

The other proof is probably "simpler". Except that you need to know more logic, to work with non-standard analysis...