A property on the unit sphere I found this lemma in:  [Furi, M., Vignoli, A.: On a property of the unit sphere in a linear normed space. Bull. Pol.
Acad. Sci. Math. 18, 333–334 (1970)]

Lemma:
  Let $S^{n-1}$ be the unit sphere of the real Euclidean space $E^n$.
  Let $\{A_1, A_2, ..., A_q\}$ be a finite family of closed sets with the following two properties:
a) $\cup_{k=1}^q  A_k =S^{n-1}$;
b)None of the sets $A_k (k= 1, 2, ..., q)$ contains two symmetric (with respect to the origin) points, then 
$q>n$

Do you have any idea of the proof?
 A: I'm going to prove the equivalent statement that if $q\leq n$ then at least one of $A_i$ contains antipodal points.
Consider the case when $q=n$ and assume that $A_{n}$ (the last closed subset) doesn't have antipodal points. If it has then we are obviously done. Consider function
$$f:S^{n-1}\to\mathbb{R}^{n-1}$$
$$f(v)_i=d(v, A_i)$$
where $d(v,A_i)$ is the distance of $v$ from $A_i$. Note that this doesn't include the last subset. This is well defined and continuous map. By the Borsuk-Ulam theorem there is $v\in S^{n-1}$ such that $f(v)=f(-v)$. But since $\{A_i\}$ cover whole $S^{n-1}$ then $v\in A_j$ for some $j$.
Now if $j<n$ then $f(v)_j=0=f(-v)_j$, i.e. $d(v,A_j)=d(-v, A_j)=0$. Because each $A_i$ is closed then $-v\in A_j$.
On the other hand if $j=n$ then we can consider $-v$. Since $A_n$ doesn't contain antipodals then $-v\in A_k$ for some $k<n$ and thus previous "$j<n$" case applies to $-v$ and $A_k$.
Therefore for $q=n$ at least one of $A_i$ contains antipodal points. $\Box$
The case when $q<n$ you solve by artifically adding singletons to $\{A_1,\ldots, A_q\}$ and going back to $q=n$. Note that singletons cannot contain antipodal points.
All in all: if $q\leq n$ then at least one of $A_j$ contains anitpodal points. Or equivalently: if none of $A_j$ contains anitpodal points then $q>n$.
Of course the hard part of the solution is the Borsuk-Ulam theorem which is far from trivial.
