Proof for the case $n = 2$:

We follow the elegant proof given in [17] (p. 109). We first show that any continuous function $g:E^2\longrightarrow S^1$ maps a pair of antipodal points on the boundary of $E^2$ to the same point. That is there exists $z \in E^2$ such that $\vert z\vert = 1$ and $g(z) = g(-z)$. Since $E^2$ is a compact convex set, by lemma (12.4) we see that any continuous map $g:E^2\longrightarrow S^1$ has a continuous lift ${\tilde g} : E^2 \longrightarrow \mathbb{R}$. Since the real valued map

$$\theta\mapsto {\tilde g}(e^{2\pi i \theta}) - {\tilde g}(e^{-2\pi i \theta}),\quad 0\leq \theta \leq 1,$$

changes sign we see that there is a pair of antipodal points $z, -z \in S^1$ such that ${\tilde g}(z) = {\tilde g}(-z)$ and hence $g(z) = g(-z)$. Turning now to a continuous map $f:S^2\longrightarrow \mathbb{R}^2$, assume $f({\bf x}) \neq f(-{\bf x})$ for every ${\bf x} \in S^2$. We construct the continuous function $g:E^2\longrightarrow S^1$

$$g(z) = h(z)/\vert h(z)\vert$$

where

$$h(x_1, x_2) = f(x_1, x_2, \sqrt{1-x_1^2-x_2^2}) - f(-x_1, -x_2, -\sqrt{1 - x_1^2 - x_2^2}),\quad (x_1, x_2) \in E^2.$$

Since $\vert h(z)\vert = \vert h(-z)\vert$, we infer that there is no $z \in E^2$ satisfying $\vert z\vert = 1$ and $g(z) = g(-z)$ resulting in a contradiction.