Proof of the Borsuk-Ulam theorem: I'm going to prove a stronger statement than the one given in the statement of the Borsak-Ulam theorem here, which is:

Every odd (that is, antipode-preserving) map $f:S^n\to S^n$ has odd degree.

Proof: We go by induction on $n$ . Consider the pair $(S^n,A)$ where $A$ is the equatorial sphere. $f$ defines a map $$\tilde{f}:\mathbb{R}P^n\to\mathbb{R}P^n$$ . By cellular approximation, this may be assumed to take the hyperplane at infinity (the $n-1$ -cell of the standard cell structure on $\mathbb{R}P^n$ ) to itself. Since whether a map lifts to a covering depends only on its homotopy class, $f$ is homotopic to an odd map taking $A$ to itself. We may assume that $f$ is such a map.

The map $f$ gives us a morphism of the long exact sequences: $$\begin{CD} H_n(A;\Zt)@>i>> H_n(S^n;\Zt)@>j>> H_n(S^n,A;\Zt)@>\partial>> H_{n-1}(A;\Zt) @>i>> H_{n-1}(S^n,A;\Zt)\\ @Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV \\ H_n(A;\Zt)@>i>> H_n(S^n;\Zt)@>j>> H_n(S^n,A;\Zt)@>\partial>> H_{n-1}(A;\Zt) @>i>> H_{n-1}(S^n,A;\Zt)\\ \end{CD}$$

Clearly, the map $f|_A$ is odd, so by the induction hypothesis, $f|_A$ has odd degree. Note that a map has odd degree if and only if $f^*:H_n(S^n;\Zt)\to H_n(S^n,\Zt)$ is an isomorphism. Thus $$f^*:H_{n-1}(A;\Zt)\to H_{n-1}(A;\Zt)$$ is an isomorphism. By the commutativity of the diagram, the map $$f^*:H_n(S^n,A;\Zt)\to H_n(S^n,A;\Zt)$$ is not trivial. I claim it is an isomorphism. $H_n(S^n,A;\Zt)$ is generated by cycles $[R^+]$ and $[R^-]$ which are the fundamental classes of the upper and lower hemispheres, and the antipodal map exchanges these. Both of these map to the fundamental class of $A$ , $[A]\in H_{n-1}(A;\Zt)$ . By the commutativity of the diagram, $\partial(f^*([R^\pm]))=f^*(\partial([R^\pm]))=f^*([A])=[A]$ . Thus $f^*([R^+])=[R^\pm]$ and $f^*([R^-]) =[R^\mp]$ since $f$ commutes with the antipodal map. Thus $f^*$ is an isomorphism on $H_n(S^n,A;\Zt)$ . Since $H_n(A,\Zt)=0$ , by the exactness of the sequence $i:H_n(S^n;\Zt) \to H_n(S^n,A;\Zt)$ is injective, and so by the commutativity of the diagram (or equivalently by the $5$ -lemma) $f^*:H_n(S^n;\Zt)\to H_n(S^n;\Zt)$ is an isomorphism. Thus $f$ has odd degree.

The other statement of the Borsuk-Ulam theorem is:

There is no odd map $S^n\to S^{n-1}$ .

Proof: If $f$ where such a map, consider $f$ restricted to the equator $A$ of $S^n$ . This is an odd map from $S^{n-1}$ to $S^{n-1}$ and thus has odd degree. But the map $$f^*H_{n-1}(A)\to H_{n-1}(S^{n-1})$$ factors through $H_{n-1}(S^n)=0$ , and so must be zero. Thus $f|_A$ has degree 0, a contradiction.