My question regarding Egoroff's Theorem mostly pertains to the conclusion that is made, that is, how come at the end of the proof we make the restriction to a closed set $F \subseteq A$. Moreover, we already have that $\{f_n\}$ converges to $f$ uniformly on $A$ (measurable) and that $m(E \setminus A) < \frac{\epsilon}{2}$, so why make the further restriction to $F \subseteq A$?
Apologies for not including the proof, however, the proof is long and can be found everywhere.
To extend this question a bit further; even in the proof of The Bounded Convergence Theorem in Real Analysis by Royden, he calls for Egoroff's Theorem, however, he just used the set $A$ for the proof rather than the actual conclusion of Egoroff's Theorem, which is for the closed set $F \subseteq A$...
By the way, the text I am using is Real Analysis by Royden. Let me know if any clarification about the question is needed.
Sometimes it suffices to obtain uniform convergence of $\{f_{n}\}$ on a merely measurable set $A$. Sometimes a bonus property, the closedness, is prefered.
For example, if you are working with a Radon measure (if you are not familiar with this terminology, just consider the Lebesgue measure on $\mathbb{R}^{d}$), you can obtain compactness by sacrificing extra $\epsilon$ (but it doesn't harm that much, as you know). A lesson from undergraduate analysis is that uniform convergence on a compact set goes well with continuity. Thus, for example, if you had a sequence of continuous functions $\{f_{n}\}$ converging to a function $f$, then for some large enough compact subset $F$, you will have $f_{n}|_{F}$ uniformly converging to $f|_{F}$: this will give you the continuity of $f$. This is one approach to prove Luzin's theorem.