Proof outline
A full proof of the theorem would be too long to reproduce here, but the following paragraph outlines a proof omitting the difficult part. It is hoped that this will at least give some idea why the theorem might be expected to be true. Note that the boundary of D n is S n-1, the (n-1)-sphere
Suppose f : D n -> D n is a continuous function that has no fixed point. The idea is to show that this leads to a contradiction. For each x in D n, consider the straight line that passes through f(x) and x. There is only one such line, because f(x) ≠ x. Following this line from f(x) through x leads to a point on S n-1. Call this point g(x). This gives us a continuous function g : D n -> S n-1. This is a special type of continuous function known as a retraction: every point of the codomain (in this case S n-1) is a fixed point of the function. Intuitively it seems unlikely that there could be a retraction of D n onto S n-1, and in the case n = 1 it is obviously impossible because S 0 isn't even connected. For n > 1, however, proving the impossibility of the retraction is considerably more difficult. One way is to make use of homology groups: it can be shown that Hn-1(D n) is trivial while Hn-1(S n-1) is infinite cyclic. This shows that the retraction is impossible, because a retraction cannot increase the size of homology groups.
There is also an almost elementary combinatorial proof. Its main step consists in establishing Sperner's lemma in n dimensions.
Generalizations
Generalizations of the Brouwer fixed point theorem to infinite dimensions include the Schauder fixed point theorem (if C is a nonempty closed convex subset of a Banach space and f is a continuous map from C to C whose image is countably compact, then '\'f has a fixed point) and the Tychonoff fixed point theorem (if C is a nonempty compact convex subset of a locally convex topological vector space, then any continuous map f from C to C'' has a fixed point).