Countable version of Erdös-Lovasz-Faber conjecture

Question

Let $X$ be an infinite set, and let $(A_n)_{n\in\omega}$ be a collection of subsets of $X$ with the following properties:

1. $|A_m\cap A_n| \leq 1$ for $m\neq n\in \omega$, and
2. $|A_n|=\aleph_0$ for all $n\in \omega$.

We consider the following statement:

(EFL$_\omega$:) There is $f:X\to \omega$ such that for all $n\in\omega$ the restriction $f|_{A_n}:A_n\to\omega$ is a bijection.

Questions. Is (EFL$_\omega$) true? Or does (EFL$_\omega$) imply the original Erdös-Faber-Lovasz conjecture?

Answer 1

If I understand it correctly, it's false. Let $x \notin A_0 = \{1,2,\dots \}$. Then let $A_i$ all meet at $x$, and also each meet $A$ at $i$ (add extra elements as necessary; they should be irrelevant). Then $f(x) \neq f(i)$ for any $i$, so $f(x) \notin f(A)$.

This didn't work in the finite case because the sets meeting $A$ cannot exhaust $A$, as the number of such sets is strictly less than the number of elements of $A$. When working over infinite sets, this is no longer true.