If $\{A_n: n \in \mathbb{N} \}$ is a sequence of subsets of $\mathbb{R}$ and $|A_n| < \mathfrak{c}$ for all $n$. Prove $| \cup_n A_n| <\mathfrak{c}$
with $\mathfrak{c}$ the cardinality of $\mathbb{R}$.
I think we want to make a bijection from one $A_n$ to $ \cup_n A_n$. Because the first has cardinality smaller than $\mathfrak{c}$, the second must have too.
Clearly if there is some $k$ such that $|A_n|\leq|A_k|$ for all $k$, then the solution you propose works just fine. However your strategy may fail. It is consistent that there are such $A_n$'s such that $|A_n|<|A_{n+1}|$, and so $\bigcup A_n$ has cardinality greater than any single $A_n$.
Suppose that is the case, and let $\kappa_n=|A_n|$, then $\kappa=\sup\kappa_n=2^{\aleph_0}=\frak c$. We have now that $\kappa^\omega>\kappa$ by Koenig's theorem, however $\frak c^\omega=c$, which is a contradiction.