If $f : S^n \to \R^n$ is continuous and antipode-preserving, then there exists a point $x \in S^n$ such that $f(x) = 0$.

Proof

Use standard stereographic projection, we can see $N$ and $S$ charts gives exactly the same coordinate system at equator (i.e. where its embedding coordinate $(x_1, \dots, x_{n+1})$ with $x_{n+1} = 0$). This also tells the equator is a $S^{n-1}$ in $\R^n$ because $x_1^2 + \dots + x_n^2 = 1$.

By continuous and antipode-preserving, $f$ preserves such an equator (we denote $E$) to $S^{n-1}$ (with any deformation that still an antipode shape) and $f(E)$ must bound a set contains $0 \in \R^n$.

By continuous, both of two open sets of $S^n$, complement of $E$, needs to be mapped to cover the subset of $\R^n$ which bounded by $f(E)$, so there exists a point $x \in S^n$ such that $f(x) = 0$.