two-part partition of the unit Euclidean closed ball In a Greek maths forum I found the following problem:

problem:
Let $\mathcal{P}=\{A,B \}$ be a partition of the $n$-dimensional Euclidean closed ball ($n>1$),
prove that at least one of the $A$ or $B$ have diameter equal with 2

if $X$ is a subset of $\mathbb{R}^n$, the diameter of $X$, $\delta(X)$, is defined to be
$$ \delta(X):=\sup\left\{ \left\lVert {\bf x} - {\bf y} \right\rVert :{\bf x}, {\bf y} \in X  \right\} $$


The question remains unanswered, but my curiosity has triggered.
Do you know how to prove this interesting result?
 A: Let's consider the corresponding statement for the unit circle in $\mathbb{R}^2$.  Namely if the set $C=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}$ is partitioned into two sets $A$ and $B$, then at least one of these two sets has diameter $2$.  Proving this statement will prove the result in the given problem (for $n\ge 2$) since each such closed Euclidean ball contains a copy of the unit circle.
Suppose our revised statement is false, and we can partition $C$ into $A, B$, each with diameter less than $2$.  Without loss of generality we may assume $A\ne\emptyset$.  Let $P\in A$.  Then the diametrically opposite point, $Q$, is in $B$.  
Claim: There is an open neighborhood $N$ of $P$ in $C$ such that $N\subseteq A$.  If the claim were false, then we could find a sequence of points in $B$ converging to $A$. But then looking at the distance of points in this sequence to $Q$, we would obtain that $B$ has diameter $2$, a contradiction.  So such an open neighborhood of $P$ exists in $A$.
From the claim, we conclude that $A$ is open in $C$.  But similarly, $B$ is open in $C$.  Since $C$ is connected, one of $A$ and $B$ must be empty.  But then certainly the diameter of the nonempty one would be $2$.
