Here are two related problems I have encountered several times before:
$i)$ Show that there can only be countably many pairwise disjoint crosses in the plane.
$ii)$ Show that there can only be countably many pairwise disjoint figure eights in the plane.
A cross is defined as the union of two line segments intersecting in their interiors. A figure eight is defined as a loop $\gamma:[0,1]\to \mathbb R^2$ with exactly one internal self-intersection. These have been answered on this site before, e.g. here and here. I am interested in some generalisations to these problems:
What is the maximal cardinality of a set of pairwise disjoint objects in the plane, if the objects are the union of:
Q1: Three line segments sharing a common end point and non-overlapping (i.e. pairwise non-parallel).
Q2: Four continuous paths with a common end point, and otherwise non-intersecting and non-self-intersecting.
Q3: Three continuous paths with a common end point, and otherwise non-intersecting and non-self-intersecting.
The objects in Q1 are three-armed rigid crosses (like the Mercedes logo). In Q2 and Q3 we have four- and three-armed wobbly starfish shapes, as it were. I strongly suspect they are all countable. If we can prove it for Q3, then Q1-2 are also countable immediately. If needed, you can assume additional smoothness to the paths, to maybe employ Taylor approximations or something.
I dont' think it is necessary to assume the paths are non-intersecting, as long they do not completely overlap. The important bit is that there is some point with three or four different paths going out from it locally.
The usual proofs for $i)$ and $ii)$ go something like this (paraphrased from answers on this site):
$i)$: Let $C_\varepsilon$ be the subset of crosses with shortest arm length $\ge\varepsilon$. The centers of two crosses in $C_\varepsilon$ can only be so-and-so close to each other, so $C_\varepsilon$ is countable. The whole collection of crosses is a countable union $\bigcup_{n\in\mathbb N}C_{1/n}$, so it is countable.
Note that I think this proof is somewhat lacking in how it determines the minimum distance of crosses. In particular, I think it is necessary to consider the angle of the cross as well, but it doesn't seem hard to fix. Given two crosses in $C_\varepsilon$ that additionally have their smaller angles $\ge\varphi$, I am sure one can find a lower bound for the distance between their centers with some elementary geometry. But if the angles are not controlled, then two crosses in $C_\varepsilon$ can be arbitrarily close.
$ii)$: Assign two rational points to each figure eight, one in the interior of each of the two bounded regions it makes. Since two figure eights can't be assigned the same pair of rational points without intersecting, we have an injection into $\mathbb Q^2\times \mathbb Q^2$.
Now, the proofs of $i)$ and $ii)$ are completely different in flavour. One is simply geometric and focuses on the intersection point. The other requires the closed loops of the figure eight to do the rational point trick. However, I think that the "real" problem (in some sense) with the figure eights is not their loops but their self-intersection.
In this vein, the objects in Q2 can be seen as open-ended figure eights. I think there should be a proof using the idea of minimal distances between crosses but accounting for the non-rigid shapes of the paths in some way. I think Q1 can be proven very similarly to $i)$ by looking at some measure of size for the objects. But I am most interested in Q3 of course, since that implies the others.
