Let $G$ be an abelian group and $| S |$ is infinite with $S\subset G$.
Then $\langle S\rangle = \bigcup \langle T\rangle$ where $T \subset S$ and $T$ is finite.
My attempt is below.
$\langle S \rangle \supset \bigcup\langle T\rangle$ is clear, since every $T$ is contained in $S$ and then each $\langle T \rangle$ is contained in $\langle S\rangle $.
Remains to show is $\langle S \rangle \subset \bigcup \langle T\rangle$.
Take any $ x \in \langle S\rangle$.
Case I. $ x \in S$. In this case, there is $ T' $ containing $x$. so it is clear.
Case II. $ x \notin S $. In this case, there are $a, b \in \langle S \rangle $ such that $ ab=x $. and then there is $ T'' $ containing $ a $ and $ b$. Then $\langle T'' \rangle$ contains $ x $.
But I'm not sure the case II. Could I say there are such $ a $ and $ b $ ? Does $ x \in \langle S \rangle$ guarantee it?