Let $f:\mathbb{R}\to [-\infty,\infty]$ continuous function.
(a) Let $\Omega=\left\{E:f^{-1}(E) \text{ is a Borel set } \right\}$. Show that $\Omega$ is $\sigma$-algebra. I already proved this. :)
(b) Let $B$ be a Borel set. Show that $f^{-1}(B)$ is a Borel set.
For (b) I have this. $f$ continuous then $f$ is measurable, because $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set. It is correct?...
It appears that you are assuming what you are supposed to prove. The purpose of this exercise is to prove that a continuous function is Borel measurable. The correct argument is to use a): the sigma algebra in a) contains all open sets because $f^{-1}(U)$ is open for any open set $U$; if a sigma algebra contains all open sets it contains all Borel sets because Borel sigma algebra is the smallest sigma algebra containing all open sets.