Suppose there is no fixed x, and proceed along the lines of the 2-dimensional proof. For each x, draw a ray starting at f(x), passing through x, and exiting the sphere at g(x). Since f(x) is continuous, g(x) is continuous. Note that g fixes every x in the sphere, i.e. the boundary of the disk. Run a linear homotopy from x to g(x). Technically, you should prove this is continuous, but I think it's pretty intuitive. Now the sphere is a retract of the disk. This implies a trivial relative homology h(Dn/Sn-1), and that contradicts an earlier theorem. Therefore there is some x such that f(x) = x.
As a corollary, a continuous function f from a set S into itself, where S is homeomorphic to a closed disk, exhibits a fixed point.
Note that this theorem fails in infinite dimensions. Consider the ball of radius 12 in Ej. Let f(x) shift the coordinates of x, divide through by 2, and place a 1 in the first coordinate. Thus f({8,6,0,4}) = {1,4,3,0,2}. Verify that f is continuous from the ball into itself, and there is no fixed point.