Theorem.
Let B={x∈R^{n} : x≤1} be the closed unit ball in R^{n}. Any continuous function f:B→B has a fixed point a=f(a).
Notes
 Shape

If D is homeomorphic to B, then Brouwer's theorem also applies to D, and any g:D→D also has some fixed point. This includes any closed balls, squares, rectangles, cubes, boxes, hypercubes, polygons without holes, oranges, tomatoes, triangles, simplices, etc., all of them closed.
 Compactness (i)

The theorem doesn't apply to R^{n}. For instance, any nonzero translation T_{v}(x)=x+v has no fixed point.
 Compactness (ii)

The theorem doesn't apply to the open unit ball B={x∈R^{n} : x<1}. For instance, f(x) = x/2 + (1/2,0,...,0) has no fixed point on B  its only fixed point is at (1,0,...,0)∉B.
 Simple connectedness

The theorem doesn't apply to torii. These have various "rotations" with no fixed points. For instance, any nonzero rotation of S^{1}={x∈R^{2} : x=1} has no fixed point. So any topological space T=S^{1}×U, no matter what U is, is a counterexample.
Brouwer's theorem is truly remarkable. It requires nexttonothing of f (just continuity), and gives a great property of f! So it is probably best regarded as a useful property of topological spaces homeomorphic to closed balls.
It is also nonconstructive. This is of necessity: what could we possibly know about f? But it means we don't even know if the fixed point of f is on the boundary of B or in its interior: it can be anywhere. The theorem is just too generic to yield a construction.
This result must have been deeply unsatisfying to Brouwer. Later, Brouwer founded the constructivist discipline of intuitionism. Constructivists reject Brouwer's theorem(!), although they will accept any particular case of a given function f and its fixed point(s).
Sperner's lemma (which see) yields a nice combinatorical proof of Brouwer's fixed point theorem. There are also proofs from algebraic topology, of course.