My initial idea was to decompose $f$ using North/South stereographic projections—the natural atlas for $S^n$. This approach not quite work here: the N/S maps send the projection point to nowhere (or must change codomain to one-point compactification), making any function decomposed this way necessarily discontinuous.

I then considered pulling the projection point outward along the N-S axis, forcing projections into finite regions to maintain continuity. This insight—that continuity constraints force us to work in bounded regions—became crucial to my understanding.

Following this idea to examine the N/S map decomposition, after compressing infinity back to $\mathbb{R}^n$, the continuity requirement also necessitates pulling back the neighborhood, leading to the same conclusion as the outward projection: the region must be finite.

At this moment, I learn the relations between different description of Borsuk-Ulam (Some applications of the Borsuk-Ulam Theorem).

My proof strategy ended up opposite to the classical development. I first established a lemma:

Lemma: Let $f : S^n \to \mathbb{R}^n$ be continuous and antipode-preserving. Then there exists $x \in S^n$ such that $f(x) = 0$.

The details can be found here, the key is $f(E) : S^{n-1} \to S^{n-1}$ bound an open set of $\mathbb{R}^n$ that contains $0$ and is the image of rest two open sets on $S^n$. Hence, $f(x) = 0$ for some $x$

For the main theorem, define $g(x) = f(x) - f(-x)$. Two key observations:

1. $g$ is continuous
2. $g$ is antipode-preserving: $g(-x) = f(-x) - f(x) = -g(x)$

Applying the lemma: there exists $x$ such that $g(x) = 0$, which means $f(x) - f(-x) = 0$, hence $f(x) = f(-x)$. Proof complete.