Is it possible to cut the unit disk in $5$ "small" parts? 
Let $D = \{(x,y) \in \Bbb R^2 \mid x^2+y^2 \leq 1\}$ be the unit disk. Is it possible to find five subsets $A_1, \dots, A_5 \subset D$ such that they cover $D$ and they all have diameter at most $1$?

My conditions just mean $$D = \bigcup\limits_{i=1}^5 A_i \qquad\text{and}\qquad
\mathrm{diam}(A_i) := \sup\limits_{x,y \in A_i} \|x-y\|_2 \leq 1,
\;\;\forall i \in \{1,\dots,5\}.$$
Of course, this is possible with $6$ pieces, namely $$A_i = \{re^{ia}    \;\mid\;    0≤r≤1,\; 2\pi (i-1) /6 ≤ a ≤ 2\pi i/6\}$$
But I don't think that this is possible with only $5$ pieces (even with non-measurable subsets), but I don't see any simple argument.
Thank you for your help! 

I will follow the idea of Daniel Fischer.

If $B_1:=A_1\cap \partial D$ can be covered by sector of central angle $t$, then $t_0=\frac{\pi}{3}$ is greatest value.

Proof. If not, then we have $x,\ y\in B_1$ s.t. $\angle xOy >\frac{\pi}{3}$, where $O$ is origin. Then $|x-y|>1$. It is a contradiction since ${\rm diam}\ A_1\leq 1$.
That is, five $A_i$ covers arc of length at most $5\pi/3$.
