If for some collection of open sets $\cup\_{i\in I} (A\_i \cup^\* -A\_i)=\mathbb S^d$, then is there an $x\in \mathbb S^d$ and $i\_1,\ldots i\_d\in I$ for which $x\in A\_{i\_1}\cap \ldots \cap A\_{i\_d}$?


Important that the question implicitly requires $A\_i\cap -A\_i=\emptyset$, as $\cup^\*$ denotes disjoint union.
Also note that I want $x$ to be in $A\_i$ and not in $-A\_i$, otherwise a suitable version of this statement would follow from Ky Fan's theorem (which implies it even in this form for $d\le 2$).


Of course this problem is very similar to the Lyusternik-Shnirelman variant of the Borsuk-Ulam theorem, so I'm sure this has been studied, but I could not find it anywhere, what of course makes it less likely to hold, which I nonetheless believe.



If $\cup (A_i \cup -A_i)=\mathbb S^d$, then is there an $x$ in $d$ sets?
Unfortunately, I think your conjecture is wrong: Embedd $S^3 \subset \mathbb C^2$ in the standard way. Consider the three sectors
$$U\_1 = \{z \in \mathbb C \ | \ z = re^{i\varphi}, r > 0, \varphi \in (0, 2\pi / 3)\}$$
$$U\_2 = \{z \in \mathbb C \ | \ z = re^{i\varphi}, r > 0, \varphi \in (2\pi / 3, 4\pi/3)\}$$
$$U\_3 = \{z \in \mathbb C \ | \ z = re^{i\varphi}, r > 0, \varphi \in (4\pi/3,2\pi )\}$$
and define
$$A\_i = \{(z\_1,z\_2) \in S^3 \ | \ z\_1 \in U\_i\}, i = 1,2,3$$
$$A\_{3+i} = \{(z\_1,z\_2) \in S^3 \ | \ z\_2 \in U\_i\}, i = 1,2,3$$
Let us check that the six sets $A\_1, A\_2, A\_3, A\_4, A\_5, A\_6$ satisfy the conditions: They are obviously open. For each $(z\_1, z\_2) \in S^3$, one of $z\_1,z\_2$ is nonzero, say $z\_1$. Then either $z\_1$ or $-z\_1$ lies in one of the sectors $U\_1, U\_2, U\_3$. Hence the six sets cover the sphere. It is also clear that $A\_i \cap (-A\_i) = \emptyset$ for all $i$. 


But now, finnally, the three sets $A\_1, A\_2, A\_3$ and $A\_4, A\_5, A\_6$ are pairwise disjoint, respectively. Hence there is no point which lies in $3$ of the sets $A\_i$.


