Here's a way of producing *any* compact subset $X$ of $S^n$ as the intersection of the boundaries of two convex bodies in $\mathbb R^{n+1}$. The first convex body is simply the unit ball $B^{n+1}$, with boundary $\partial B^{n+1}=S^n$.

Te second convex body, call it $K$, is constructed as follows.

Consider a function $f:S^n\to \mathbb R$ such that $f^{-1}(0)=X$.
Such functions exist in great abundance, see e.g. the answers to this question.
By carefully selecting $f$ (i.e. by taking it to be small, and with small first and second derivatives), we can make sure that
$$
K:=\{x\in\mathbb R^{n+1}:\|x\|\le 1+f(x/\|x\|)\}
$$
is convex.

It is then clear, by construction, that $\partial B^{n+1}\cap \partial K = X$.

If $f$ admits both positive and negative values (which can be arranged iff the complement of $X$ in $S^n$ is disconnected), then neither of $B^{n+1}$ or $K$ is contained in the other one.