continuity and open sets

[gefest_md]

Good evening. Here are two citations from a book about basic principles of analysis:

Цитата:

Exercise 1. A function $f\colon\mathbb{R}\to\mathbb{R}$ is continuous if and only if the preimage of every open set is open.

Цитата:

Exercise 2. If $f\colon\mathbb{R}\to\mathbb{R}$ is continuous on a compact set $S$, then $f(S)$ is compact.

I know that exercise 1 can help to prove exercise 2. I am starting (perhaps hastily) by defining $g\colon S\to\mathbb{R},\ g(x)=f(x)$. Then what allows to apply exercise 1 to function $g$ ($f\colon\mathbb{R}\to\mathbb{R}$ but $g\colon S\to\mathbb{R}$)? I am even inclined to reprove exercise 1 for $g$.

[Oleg Zubelevich]

cover $f(S)$ with an open covering, look at the preimage of that covering

[gefest_md]

I am going to look at some collection $\left\{U_\alpha\right\}$. But as an amateur, I can not understand the usefulness of Exercise 1, with $f\colon\mathbb{R}\to\mathbb{R}$ in it, when I had to infer that each of $\left\{f^{-1}\left(U_\alpha\right)\right\}$ is also open. With Exercise 1 in hand, why Exercise 2 can be addressed on restricted domain?

[Oleg Zubelevich]

Actually the both assertions hold true not only for the case $\mathbb{R}$ but for general topological spaces. I believe you should use a proper course of analysis L Schwatz, for example