example of uncountable family of subsets of a countable set with finite intersections

We wish to give an answer to the following:

Problem. Assume, that $X$ is a countable set. Is there a family ${\{X_i\}}_{i\in I}$ of subsets of $X$ such that $I$ is an uncountable set, but for any $i\ne j\in I$ the intersection $X_i\cap X_j$ is finite?

Example. Let $x\in [1,2)$ be a real number. Express $x$ using digits

$$x=1.x_1{x}_2{x}_3{x}_4\mathrm{\cdots }=1+\sum _{i=1}^{\mathrm{\infty }}{x}_i\cdot {10}^{-i}$$

where each $x_i\in \{0,1,2,3,4,5,6,7,8,9\}$. With $x$ we associate the following natural numbers

$${\beta }_n(x)=1x_1{x}_2{x}_3\mathrm{\cdots }{x}_{n-1}{x}_n={10}^{n+1}+\sum _{i=1}^n{x}_i\cdot {10}^{n-i+1}.$$

Now define $A:[1,2)\to \mathrm{P}(\mathbb{N})$ (here $\mathrm{P}(X)$ stands for ,,the power set of $X$”) by

$$A(x)=\{{\beta }_1(x),{\beta }_2(x),{\beta }_3(x),\mathrm{\dots }\}.$$

$A$ is injective. Indeed, note that for any $x,y\in [1,2)$ if ${\beta }_i(x)={\beta }_j(y)$, then $i=j$ (this is because equal $\beta$ numbers have equal ,,length” and this is because each $\beta$ has $1$ at the begining, zeros are not the problem). Therefore, if $A(x)=A(y)$ for some $x,y$, then it follows, that ${\beta }_i(x)={\beta }_i(y)$ for each $i$, but this implies that corresponding digits of $x$ and $y$ are equal. Thus $x=y$.

This shows, that ${\{A(x)\}}_{x\in [1,2)}$ is an uncountable family of subsets of $\mathbb{N}$. Now in order to prove that $A(x)\cap A(y)$ is finite whenever $x\ne y$ it is enough to show that we can uniquely reconstruct $x$ from any infinite sequence of numbers from $A(x)$. This can be proved by using similar techniques as before and we leave it as a simple exercise.