Assume that $\mathcal{A}$ is a countable infinite set of infinite subsets of $\mathbb{N}$ such that $A \cap B$ is finite for all $A \neq B \in \mathcal{A}$. Prove that there is an infinite set $X \subset \mathbb{N}$ such that $X \cap A$ is finite for all $A \in \mathcal{A}$.
I want to construct the set $X$ in the following way,
$\forall n$, $x_n \in X$, is such that $x_n=min \left ( (A_n \setminus \bigcup_{i <n} A_i) \setminus \{x_0,x_1, \dots, x_{n-1} \} \right )$.
Example, $x_0=min(A_0)$
$x_1= min(A_1 \setminus A_0)\setminus \{x_0\}$
Then $X= \bigcup \{ x_i | i \in \omega \}$. I think this will work but I am having a difficult time with the details.
How can I construct the $x_i$ using recursion?