Lots of sets, finite intersections

This is a sketch. I assume $\mathbb{N} = \{0,1,2,3,\ldots\}$ , although there's no difference.

First, we know that $|\mathbb{N}| = |\mathbb{N}^2|$ , so we might as well pick subsets of $\mathbb{N}^2$ and use a bijection from $\mathbb{N}^2$ to $\mathbb{N}$ to translate it.

Consider the first quadrant of the Euclidean plane. Partition it into unit squares. With each unit square, we can denote a pair of naturals $(x,y)$ representing its lower-left corner (the corner closest to the origin). Thus we might as well pick sets of these squares and use the bijection to get subsets of $\mathbb{N}^2$ .

Now, for each positive real $r$ , consider the line $y = rx$ . This passes through some of the squares. (Assume that a line passes through a square that shares any point with it, including corners; this doesn't really matter.) Now we have an uncountable family of sets of squares; it just remains to show that these sets have finite intersections.

Consider two lines $y = r_1 x$ , $y = r_2 x$ . The two lines will eventually be too far apart to share any squares in common; in other words, there exists some $R$ such that if we only consider the portions of the lines lying outside $x^2+y^2 = R^2$ , they are separated by at least $4$ units everywhere, and will thus never share any square. Thus all squares they share in common lie inside the circle (or on the circle), in which there are only a finite number of them. This proves the claim.

We claim that such subsets exist. Let $a = (a_1, a_2, a_3, \dots)$ be an infinite binary sequence, so $a_i \in \{0,1\}$ for all $i$ . (By Cantor's diagonal argument, the number of infinite binary sequences is uncountable.)

To each such sequence $a$ , we associate a set of positive integers $S_a$ as follows: Let $S_a$ be the set of all positive integers of the form $p_1^{a_1} p_2^{a_2} \dotsm p_i^{a_i},$ where $p_i$ is the $i^{\text{th}}$ prime.

Let $b$ be another infinite binary sequence. Then there exists a $j$ such that $a_j \neq b_j$ , so the only elements that $S_a$ and $S_b$ can have in common are the elements of the form $p_1^{a_1} p_2^{a_2} \dotsm p_i^{a_i},$ where $i < j$ . In particular, $S_a$ and $S_b$ have finite intersection.

Here's a construction obtained with the help of some basic analysis. Let $\mathbb{Q}$ be the set of all rational numbers. We know that $\mathbb{Q}$ is countable, so let $f: \mathbb{Q} \to \mathbb{N}$ be a bijection. For every irrational number a, let $S_a$ be a sequence of rational numbers converging to a. Since every convergent sequence has a unique limit, for every distinct irrational numbers a and b, the intersection $S_a \cap S_b$ must be finite. Therefore $\{f(S_a): a \in \mathbb{R} \setminus \mathbb{Q}\}$ is an uncountable family of sets of natural numbers where every distinct pair of sets has finite intersection.