The problem/puzzle is:
Find two convex sets in Euclidean space, $A, B\subseteq\mathbb{R}^d$, such that the number of connected components of $\mathbb{R}^d\setminus (A\cup B)$ is the maximum attainable.
I am interested in both proofs of upper-bounds and sets that achieve the upper-bound.
I have found two convex sets in $\mathbb{R}^2$ that create $5$ such components, and conjecture that the maximum attainable in $d$-dimensions is $2d+1$.
Example:
the lines $x=0$, $y=0$ in $\mathbb{R}^2$ yield $4$ connected components, each of the quadrants.
Here's how to obtain $2d+1$ components:
For $d=1$, let $A_1=[-1,0)$ and $B_1=(0,1]$
Assume we have found $A_{d-1}$, $B_{d-1}$ that produce $2d-1$ components in $d-1$ dimensions. Let $$\begin{align}A_d&=\{\,(x_1,\ldots,x_d):0<x_d\le1\,\}\cup A_{d-1}\times\{0\},\\B_d&=\{\,(x_1,\ldots,x_d):-1\le x_d<0\,\}\cup B_{d-1}\times\{0\}.\end{align}$$ Then $\mathbb R^d\setminus(A_d\cup B_d)$ has the two components "$x_d>1$" and "$x_d<-1$" as well as the $2d-1$ components from the lower dimensional solution in the "$x_d=0$ hyperplane.
We can show that $2d+1$ is optimal by induction on $d$: Assume we have for some $d\ge 2$ two convex subsets $A,B\subseteq \mathbb R^d$ such that $\mathbb R^d\setminus (A\cup B)$ has $m$ components $C_1,\ldots,C_m$. We shall exhibit a hyperplane $H\cong \mathbb R^{d-1}$ such that at least $m-2$ of the $C_i$ intersect $H$. Then $A\cap H$ and $B\cap H$ are convex subsets such that the complement of their union has at least $m-2$ components.
Lemma. Let $U$ be an open halfspace with boundary $H$ and such that $U\cap B=\emptyset$, $H\cap (A\setminus B)\ne\emptyset$. Then among the $C_i$ with $C_i\cap U\ne\emptyset$ there is at most one with $C_i\cap H=\emptyset$.
Proof. Let $a\in H\cap (A\setminus B)$ and $c,c'\in U$ belong to distinct components of the complement. Then in the plane $\pi$ determined by $a,c,c'$ we have the situation of the following image (where the black line is $H\cap \pi$):
In order to prevent that $c,c'$ are connected via the green path, there must be a point $a'\in A$ on the top line segment (parallel to $H$). Then $A\cap \pi$ is confined to the blue shaded area. We see that the components of $c$ and $c'$ both intersect $H\cap \pi$ in an interval. But at most one of these can be $\subseteq B$ because they are separated by $a\notin B$. $_\square$
Corollary. If $H$ is a hyperplane such that $A$ is contained in one and $B$ contained in the other closed halfspace detemined by $H$ then $H$ intersects at least $m-2$ of the $C_i$.
Proof. The lemma as stated and with $A,B$ swapped shows that each of the two halfspaces can contain at most one component not intersecting $H$. $_\square$
Proposition. Let $\ell$ be a line that intersects at least three of the $C_i$. Then a hyperplane intersecting at least $m-2$ of the $C_i$ exists.
Proof. Let $c,c',c''$ be points on $\ell$ belonging to different $C_i$, with $c'$ between $c$ and $c''$. Then wlog. the open line segment $cc'$ intersects $A$ in (at least) a point $a$ and the open line segment $c'c'$ intersects $B$ in (at least) a point $b$.
There exists halfspace $U_A$ with $c\in\partial U_A$ and $A\subseteq \overline{U_A}$. Similarly we have $U_A'$ with $c'\in\partial U_A'$ and $A\subseteq \overline{U_A'}$; and $U_B'$ with $c'\in\partial U_B'$ and $B\subseteq \overline{U_B'}$; and $U_B''$ with $c''\in\partial U_B''$ and $B\subseteq \overline{U_B''}$.
Case 1: We have $\ell\subset \partial U_A'$. Assume $B$ intersects $U_A'$. Then in the plane through $\ell$ and a point $b'\in B\cap U_A'$ we have this situation:
Here $A$ is confined to the blue area and $B$ to the blue plus red area, by which $c$ and $c'$ are conected. We conclude that $B\cap U_A'=\emptyset$. But then we are in the situation of the corollary and are done.
Case 2: We have $\ell\subset\partial U_B'$. This is symmetric to case 1.
Case 3: We have that $\partial U_A', \partial U_B'$ intersect $\ell$ transversally and are equal. Then $A$ and $B$ must be back to back, i.e., we are once more in the situation of the corollary.
Case 4: We have that $\partial U_A', \partial U_B'$ intersect $\ell$ transversally and are different. This implies that $U_A'\cap U_B'\ne \emptyset$. We may assume also that $\partial U_A$ and $\partial U_B''$ intersect $\ell $ transversally as otherwise we could have just as well taken $U_A'=U_A$ or $U_B=U_B''$. Then in plane $\pi$ through $\ell$ and a point $\in U_A'\cap U_B'$ we are in this situation:
[to be continued]