When the author considers the set $F'$ some moments are unclear.
1) Is it true that $A_{\varepsilon/3}$ contains $\bigcup\limits_{n\geq N}E_n$ ? If yes, how to prove it?
2) Why the author takes the condition $\sum \limits_{n\geq N}2^{-n}<\varepsilon/3$? What is the consequence of this inequality?
Would be very grateful if anyone can answer my questions.
1) No, not necessarily, and this is not needed. 2) To get the desired $m(E - F_{\epsilon}) \leq \epsilon$ (see below).
Found the book. Both this proof and the one I know are based on a) Egorov's theorem (Theorem 4.4 there) and b) the fact that for any measurable set $F$ and any $\delta > 0$ there is a closed measurable subset $G$ of $F$ such that $m(F - G) < \delta$ (part of Theorem 3.4 there).
In the proof above: $m(E - A_{\epsilon/3}) \leq \epsilon/3$; $A_{\epsilon/3} - F' \subseteq \bigcup_{n \geq N}E_n$ and therefore (here is the answer to your second question!) $m(A_{\epsilon/3} - F') \leq \epsilon/3$; finally, $m(F' - F_{\epsilon}) \leq \epsilon/3$. Together these give $m(E - F_{\epsilon}) \leq \epsilon$.
An alternative way is to prove the theorem for step functions first (in full strength, using just the Theorem 3.4), and, for general case, to use Egorov's theorem (which gives $F \subseteq E$ with $m(E - F) \leq \epsilon/2$ and $f_n \to f$ uniformly on $F$) and the case proven already (which gives closed $F_n \subseteq F$ with $m(F - F_n) \leq \epsilon / 2^{n + 1}$ and $f_n$ continuous on $F_n$; finally we're done with $F_{\epsilon} = \bigcap_{n = 1}^{\infty} F_n$).