Let $S$ be a set with $\lvert S\rvert=\lvert\mathbb{R}\rvert$. Suppose it has subsets $S\_x$ indexed by $x\in \mathbb{R}$ with $\lvert S\_x\rvert=\lvert\mathbb{R}\lvert$ for each $x\in \mathbb{R}$. Suppose that


* for any $s\in S$ we have $\lvert\{x\in\mathbb{R}\mathrel\vert s\in S\_x\}\rvert=2$
* for any $x\neq y\in \mathbb{R}$ we have $\lvert S\_x\cap S\_y\rvert=3$.


Consider functions $f:S\to \coprod\_{x\in\mathbb{R}}S\_x$ such that $f\circ \pi=\mathrm{id}$ where $\pi:\coprod\_{x\in\mathbb{R}}S\_x\to S$ is the projection. They are indexed by $2^{\mathbb{R}}$ because each $s\in S$ can go in $2$ subsets. Is there $f$ such that $\lvert f(S)\cap S\_x\rvert<\infty$ for each $x\in \mathbb{R}$?



Include each point of continuum in a subset so that each subset gets finitely many points
Let $M : = \mathbb R^2 \setminus\{(x, y): x^2 + y^2 \leq 1\}$, $\Delta := \{(x, x) \in \mathbb R^2\}$, and let $h$ be any bijection from $\mathbb R$ to the circle $\{(x, y) \in \mathbb R^2: x^2 + y^2 = 1\}$.


Define $S := (\mathbb{R}^2 \setminus \Delta) \sqcup M \sqcup (\mathbb R/{\sim})$, where $x \sim y$ if and only if $h(x)$ and $h(y)$ are diametrically opposite points. For $\alpha \in \mathbb R$, let $S\_\alpha$ consist of the points of the form $$L\_\alpha := \{(y, \alpha) : y \in \mathbb R \setminus \{\alpha\}\} \cup\{(\alpha, y) : y \in \mathbb R \setminus \{\alpha\}\} \subset \mathbb R^2 \setminus \Delta $$ (i.e. the vertical and horizontal lines going through $(\alpha, \alpha)$) as well as the tangent line to $h(\alpha)$ in $M$, as well as the image of $\alpha$ under the $\mathbb R \to \mathbb R/{\sim}$ map.


First of all, note that this satisfies the properties you described. For $s = (a, b) \in \mathbb R^2 \setminus \Delta$, $s$ lies in $S\_a$ and $S\_b$; for a point $s \in M$, $s$ lies in the two sets $S\_\alpha$ for which the line going through $s$ and $h(\alpha)$ is tangent to the circle, and of course every point in $\mathbb R /{\sim}$ also has two preimages.


It's also clear that any two $S\_\alpha$, $S\_\beta$ intersect at three points: two points in $\mathbb R^2 \setminus \Delta$ and one either in $\mathbb R/{\sim}$ or in $M$ depending on whether $h(\alpha)$, $h(\beta)$ are diametrically opposite points on the circle or not.


Lastly, for this example, the described function $f$ cannot exist. Indeed, suppose $f(S) \cap S\_x$ is finite for every $x$. Then in particular you have finitely many points mapping to each $L\_x$, and yet the union
$$\bigcup\_{x \in \mathbb R} L\_x \cap f(\mathbb R^2 \setminus \Delta)$$ has to be all of $\mathbb R^2\setminus \Delta$.
To paraphrase, you have a finite set of points on every vertical and horizontal line in $\mathbb R^2$ and their union is all of $\mathbb R^2$ — that cannot happen (indeed, let $V\_a$ be the set of $x$ coordinates of the finite set of points on the line $y = a$ and choose $s$ which isn't in $V\_n$ for any $n \in \mathbb Z$; then to fill all of $\mathbb R^2$, the horizontal line $x = s$ would have to contain all the points $(s, n)$ for all $n \in \mathbb Z$).


