Can I decompose a compact set in a finite number of convex set? My problem is in a finite-dimensional space. I look at $\mathcal{X}$ the support of a function $f$, that is continuous and has bounded support.
\begin{eqnarray}
\mathcal{X}_o & = & \{x \in \Omega, f(x)>0 \} \\
\mathcal{X} & = & \bar{\mathcal{X}}_o \\
 \lVert x \lVert < M& &  \; \forall x\in \mathcal{X}
\end{eqnarray}
I already performed many tricks using the Borel-Lebesgue property (Given an union of open set covering a compact set $\mathcal{X} \subset \cup_i B_i$, we can cover $\mathcal{X}$ with a finite number of open sets $\mathcal{X} \subset \cup_{i \in J} B_i$ with $|J|$ finite) [or Heine–Borel theorem]. 
For one point, I cover $\mathcal{X}$ by convex open sets $B_i$ (open balls). I would need $B_i \cap \mathcal{X}$ to be convex for all $i$. Can we prove the existence of such a decomposition ? 
PS : seems provable to me. It seems that such a cover can be built by balls $B_i$ small enough. Or, if I can decompose $\mathcal{X}$ into a finite number of compact convex sets $\cup_{i=1}^mC_i = \mathcal{X}$, then I would just do the Borel-Lebesgue trick on each of those $C_i$. Otherwise, if somebody has a counter example of compact set that cannot be covered by a finite number of convex sets, it would solve the problem.
Thank you very much
Arnaud

As Daniel Fischer said (in a comment, unfortunately), the answer is negative. One counterexample is a round annulus $1\le |x|\le 2$ in dimension $2$ (or higher).

Any convex subset of that can meet at most one point of the inner bounding circle. 

