Proposition

Whenever $S^2$ is expressed as the union of three closed sets $A_1, A_2$, and $A_3$, then at least one of these sets must contain a pair of antipodal points $\{ x, -x \}$.

Solution

Since $S^2$ is nonempty, not all $A_1, A_2, A_3$ are empty.

If only one of them is nonempty, then it contains $(1, 0, 0)$ and $(-1, 0, 0)$, so we are done. Suppose that at least two of them are nonempty. Without loss of generality, suppose that $A_1 \ne \emptyset$ and $A_2 \ne \emptyset$.

For each $i = 1, 2$, define $d_i(x) = \inf \{ d(x, y) \mid y \in A_i \}$. $d_1, d_2$ are well defined because we assumed that $A_1$ and $A_2$ are both nonempty.

Let $g: S^2 \rightarrow \mathbb{R}^2$ be defined such that $g(x) = (d_1(x), d_2(x))$. By the Borsuk-Ulam theorem, there exists an $x \in S^2$ such that $d_1(x) = d_1(-x)$, and $d_2(x) = d_2(-x)$.

Suppose that $d_1(x) = d_1(-x) = 0$. Then $x$ and $-x$ are limit points of $A_1$. Since $A_1$ is closed, $A_1$ contains both $x$ and $-x$.
Suppose that $d_2(x) = d_2(-x) = 0$. Then $x$ and $-x$ are limit points of $A_2$. Since $A_2$ is closed, $A_2$ contains both $x$ and $-x$.
Suppose that $d_1(x) = d_1(-x) \ne 0$ and $d_2(x) = d_2(-x) \ne 0$. Then $x \notin A_1, -x \notin A_1, x \notin A_2$, and $-x \notin A_2$. This means $x, -x \in A_3$.