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?

Let $A$ be an infinite set. Then $A$ has a countable subset $B=\{b_0,b_1,b_2,\dots\}$.
Let $C=A\setminus B$. Let $U_i=\{b_i\}\cup C$. The $U_i$ cover $A$, but no finite subcollection of the $U_i$ covers $A$. 
