Lusternik-Schnirelmann theorem

Statement
If $$S^n$$ is the union of $$n+1$$ closed subsets, then at least one of those subsets contains a pair of antipodal points.

Proof for two dimensions
The proof in the two-dimensional case follows from the Urysohn lemma, and the Borsuk-Ulam theorem in two dimensions. The idea is as follows:

Suppose $$S^2$$ is a union of closed subsets $$F_1,F_2,F_3$$. Let $$a$$ denote the antipode map. Suppose that $$F_1 \cap a(F_1)$$ is empty and $$F_2 \cap a(F_2)$$ is empty. Then, by the Urysohn lemma, and the fact that $$S^2$$ is a normal space, construct a continuous function $$g_1: S^2 \to I$$ that takes the value 0 on $$F_1$$ and 1 on $$F_2$$. Analogously, construct a continuous function $$g_2:S^2 \to I$$ that takes the value 0 on $$-F_1$$ and 1 on $$-F_2$$. Then the function $$f = (g_1,g_2)$$ is a continuous function from $$S^2$$ to $$\R^2$$.

By the Borsuk-Ulam theorem, there exists $$x \in S^2$$ such that $$f(x) = f(a(x))$$. By assumption $$x$$ cannot be in $$F_1$$ because that would mean $$g_1(a(x)) = g_1(x) = 0$$ forcing $$a(x) \in F_1$$. Similarly $$x$$ cannot be in $$F_2$$. This forces $$x$$ to be in $$F_3$$. Similarly, $$-x$$ must also be in $$F_3$$, and we are done.