Countable exhaustions have finite refinements implies finite? Let $S$ be a set that satisfies the following property:

For every countable family $\{U_{i}\}_{i\in \mathbb{N}}$ of subsets of $S$ for which $$S=\bigcup_{i\in \mathbb{N}}U_{i}$$there is a finite set $I\subset \mathbb{N}$ for which $$S=\bigcup_{i \in I}U_{i}$$

Does it follow that $S$ is finite?

(I’m assuming the axiom of choice here.) Suppose that $S$ is infinite; then there is an injection $f:\Bbb N\to S$. Let $A=S\setminus f[\Bbb N]$. For $n\in\Bbb N$ let $U_n=A\cup\{f(k):k<n\}$.


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*What is $\bigcup_{n\in\Bbb N}A_n$?  

*Does $\{U_n:n\in\Bbb N\}$ have a finite refinement?

