Let P be a finite set of points on the unit sphere $S^2$ such that

for every $p\in P$, there exists a closed curve $\gamma_p \subset S^2$ which has a self intersection at $p$ and passes through $-p$. Moreover, every plane passing through $p$ and the origin intersect $\gamma_p$ in at most 2 points, excluding $-p$ (see the graph below by Neil Strickland).

At the self intersection point $p$, $\gamma_p$ is orthogonal to itself.

$\gamma_{p_1} \cap \gamma_{p_2} \subset P$ for every $p_1,p_2 \in P$

If $A_i \subset S^2$ be such that $\partial A_i \subset \gamma_{p_i}$, $p_i\in P$, i=1,2, then neither $A_1 \subset A_2$ nor $A_2 \subset A_1$.

$P \subset \gamma_p$ for all $p\in P$.

Can we prove that $P$ has either one or infinite elements? It seems that more than one element leads to infinite number of elements in $P$. It is easy to see that if $P$ has more than one point, then it has at least 6 points.

Please see the picture provided by Neil Strickland for a visualization of $\gamma_p$.

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