Let $S^2$ be the surface of the unit ball in $\mathbb{R}^3$ and let $T\colon S^2\rightarrow \mathbb{R}$ be a continuous map.
How could we justify that $T$ gets a maximum $T_{\text{max}}$ and a minimum $T_{\text{min}}$ and that there is a point $x_0\in S^2$ with $T(x_0)=T(-x_0)$ ?
Could you give me a hint? I don't really have an idea.
Since $S^2$ is compact, the maximum and the minimum of $T$ must be attained somewhere.
Suppose that you never have $T(x)=T(-x)$. Then the map$$\begin{array}{rccc}f\colon&S^2&\longrightarrow&\mathbb R\\&x&\mapsto&T(x)-T(-x)\end{array}$$is continuous and it has no zeros. But this is impossible, because if $f(x)>0$, then $f(-x)<0$ and, since $S^2$ is connected, $f(S^2)$ must be connected too and therefore it is an interval.