Include each point of continuum in a subset so that each subset gets finitely many points 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 $x\in \mathbb{R}$?
 A: 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 $n \in \mathbb Z$).
