Can a symmetric, open, simply connected set in $\mathbb{R}^2$ be partitioned into two congruent sets? Can an open, "symmetric" (in a non-rigorous sense), simply connected set $S$ in $\mathbb{R}^2$ be partitioned into two congruent sets? This sounds obvious (eg. for a circle, cut it in half) but there is a technicality that I am getting at which makes me ask this question.
Here, I defined two sets to be congruent if they are related by an isometry (a continuous, distance preserving function). Take the domain to be $\mathbb{R}^2$, where the isometries are translations, rotations, and reflections. For example, the sets $(0, 1) * {0}$ and $(0, 1) * {1}$ are obviously congruent, but they are not congruent to $[0, 1] * {1}$ (as far as I am aware!) because they isometry that brings the first to the second set would not include the endpoints. This is the technicality that leads me to my question (as opposed to asking about if the set admits a decomposition into two congruent sets with disjoint interior, in which case "cut it in half" works fine.)
For example, for the open unit disc, we might guess that the two sets should be the two halves, but then the question becomes what to make of the line in the middle, and how to assign them to each set. The closest I can come up with is to exclude the origin from consideration, and have one radius lie entirely within one set and the diametrically opposite radius lie in the other set. Then, these two halves are congruent (related by a 180 degree rotation around the origin), but at the price of losing the origin. Thus, the open unit disc punctured at the origin (ie. no longer simply connected) can be partitioned into two congruent sets.
For finite sets we definitely need there to be an even number of points, but this breaks down at even the countably infinite point size.
Relatedly, for a unit disc punctured somewhere other than the origin, this shape is still symmetric but the naive method of "cut it in half through the puncture point" doesn't work because the two radii described above aren't symmetric, so there becomes a problem of how to allocate points on the radius passing through the puncture point, with no antipodal point to compensate and save us. Is the extent / quality of symmetry required in order to admit a decomposition easy to describe?
Finally, this question seems close to the Banach Tarski paradox, which says that between two sets A and B (in dimensions strictly more than two), we can decompose both sets into $k<\infty$ pieces $A = \cup A_i, B = \cup B_i$ such that $A_i$ is congruent to $B_i$. So if we partition $S$ into two arbitrary pieces $C, D$ then we can find pieces $C_i, D_i$ which are congruent and union up to $C, D$ - but because the spatial arrangement between the $C_i, C_j$ is not specified, it is possible that $C_i \cup C_j$ is not congruent to $D_i \cup D_j$ - so $C, D$ are not congruent.
 A: The open unit disk $D$ cannot be partitioned into two congruent subsets.  Suppose such a partition $D=A\cup B$ did exist; we may assume $(0,0)\in A$.  Let $f$ be an isometry of $\mathbb{R}^2$ that maps $A$ to $B$.  Rotating our sets, we may assume that $f(0,0)=(t,0)$ for some $t\neq0$.  Since every point of $A$ has distance less $1$ from $(0,0)$, every point of $B$ has distance less than $1$ from $(t,0)$.  In particular, this means that for all sufficiently small $\epsilon>0$, $(0,1-\epsilon)\in A$ and also $(0,\epsilon-1)\in A$, so $(0,1)$ and $(0,-1)$ are both in the closure $\overline{A}$.  Thus $f(0,1),f(0,-1)\in\overline{B}\subseteq\overline{D}$ are a pair of points in the closed unit disk of distance $2$ apart, so they must be antipodal points on the unit circle.  Thus the midpoint of $f(0,1)$ and $f(0,-1)$ is $(0,0)$, and so $f(0,0)=(0,0)$ since $f$ preserves midpoints.  This is a contradiction.
A: Here is a construction when $S={\mathbb R}^2$, easily generalizable to (nonempty, simply-connected, open) subsets $S$ in the plane invariant under the action of an infinite cyclic group of translations.
Define a collection of subsets of ${\mathbb R}^2$ by
$$
A_n=\{(x,y): n\le x<n+1\}, n\in {\mathbb Z}. 
$$
Set
$$
A:= \bigcup_{n\in 2 {\mathbb Z}} A_n$$
$$
B= \bigcup_{n\in 1+ 2 {\mathbb Z}} A_n$$
Then the translation $T: (x,y)\mapsto (x+1,y)$ is a congruence
$A\to B$. At the same time, clearly, ${\mathbb R}^2= A\sqcup B$.
