A conjecture on antipodes and Jordan curves on the sphere

Question

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 $S-C$.

Answer 1

Assume the contrary.

If $x\in C$, then it follows from the Schoenflies theorem $x$ is the limit of points $x_i$ in $A$ and $y_i$ in $B$. If we assume $x\in C\smallsetminus (-C)$, we can assume in addition that $x_i$ belongs to $A\smallsetminus (-C)$ and $y_i$ to $B\smallsetminus (-C)$. By the contradictory assumption, $-x_i$ belongs to $A$ and $-y_i$ to $B$. So $-x$ is in the closure of both $A$ and $B$, hence belongs to $C$.

Hence $-(C\smallsetminus (-C))\subset C$. But the inclusion $-(C\cap (-C))\subset C$ is trivial. Accordingly, $-C\subset C$. Therefore $-C=C$. Then $x\mapsto -x$ is a fixed-point-free self-homeomorphism of the closed disc $A\cup C$ (this is a homeomorphic copy of the closed disc by the Schoenflies theorem), and this is a contradiction.

Answer 2

Let's say $C$ divides $S^2$ into two disks, $D$ and $E$. Then, choosing a homeomorphism (fixing $C$) $h:D\to E$, we get an involution $t: S^2\to S^2$, which has degree $-1$. On the other hand, since $t(x) = h(x)$ (or $h(t(x)) = x$, depending on whether $x$ is in $D$ or $E$), $t(x) \neq - x$ for any $x\in S^2$. Thus we may find a homotopy from $t$ to $\mathrm{id}_{S^2}$ (straight line, normalized). This shows that the degree of $t$ is $+1$, a contradiction.