I've been thinking about this statement for a while, and I think it's true, but I'm not sure of how to prove it. The statement is
A Jordan curve $J$ that is symmetric about the origin $p$ does not pass through $p$.
I think one should be able to argue by contradiction that if $J$ passes through the origin, then $J$ has a self-intersection, which contradicts the fact $J$ is a Jordan curve, but I'm having trouble with the details.
OK: Suppose that $J\subset R^n$ is a Jordan curve invariant under the antipodal map $\phi: x\mapsto -x$ and $0\in J$. Then $\phi$ restricts to an involution $\tau$ on $J-\{0\}\cong {\mathbb R}$: $\tau\circ \tau=id$. Since $\phi$ has only one fixed point, $\tau$ has no fixed points. Now, contradiction comes from the following lemma
Lemma. Let $f: {\mathbb R}\to {\mathbb R}$ be a homeomorphism of finite order $k$, i.e. $f^k=id$. Then $f$ has at least one fixed point.
I will leave you a proof as an exercise, compare this question.