$A \subset \mathbb {R} , f$ continuous then $f(A)$ is contable union of closed.

Question

If $A \subset \mathbb{R}$ is $\mathcal{F}_{\sigma}$ (countable union of closed) and $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, show that $f(A)$ is $\mathcal{F}_{\sigma}$. \ \

I have a question with the interpretation of the problem. Given a subset $ A $ of $ \mathbb{R} $, and $ f (A) $ is continuous. I must see that $ f (A) = \cup_{i = 1}^ {n} F_i$, with $F_i $ closed, if $ A $ was compact, we will have such a statement since $ f (A) $ is compact, but when not, I don't know where to route the proof ...

Answer 1

Hint Show that $A$ is countable union of compact sets.

Hint 2 If $C$ is closed in $\mathbb R$ then $C \cap [n,n+1]$ is compact.