I got the following conjecture: Let $C$ be a Jordan curve on the sphere $S:=S^2$ and let $A$ and $B$ be the connected compnents of $S-C$. Then there is a pair of antipodes $a$ and $b$ such that $a\in A$ and $b\in B$

I managed to prove it in the case that there is a point $p\in S-C$ with its antipode on $C$ as follows. Denote by $\alpha(x)$ the antipode of $x$. Say that $p\in A$. Define for some positive number $M>>0$ a function $f$ on $S$ such that $f(x)=M. dist(x,C)$ for $x\in S-A$ and $f(x)=-dist(x,C)$ for $x\in A$. Define also $g(x)=f(x)-f(\alpha(x))$. Now, $g(p)<0$. Choose $M$ large enough such that we have a point $q\in B$ with $g(q)>0$. By a strenghtening of the Jordan curve Theorem, $A$ and $B$ are connected by paths and have $C$ as common boundary. Thus we may pick a path $P$ from $p$ to $q$ such that $P\cap C=\{\alpha(p)\}$. By the intermediate value theorem, there is $y\in P$ with $g(y)=0$. As $g(\alpha(p))\neq 0$, $y\notin C$. This implies that $y$ and $\alpha(y)$ are in different connected components of $S-C$.