Partitioning $\mathbb{R}^d$ with two convex sets 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.
 A: 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]
A: Sorry, I've realized I never provided the example for 5 connected components. Here it is in notation with respect to $\mathbb{R}\times\mathbb{R}$:
$$ A=\big((-\infty,\infty)\times (0,1]\big)\bigcup \big(\{0\}\times\{0\}\big)$$
$$B=\big((-\infty,\infty)\times[-1,0)\big)\bigcup \big(\{1\}\times\{0\}\big)$$
The union forms two parallel strips separated by an empty line with two points on it. So in terms of the problem the connected components are the outsides of either side of the strips, the two rays on the inside and the segement $(0,1)\times\{0\}$.
