Here's one that I rather like that uses equicontinuity and compactness as the engine for moving from pointwise to uniform convergence. For the sake of clarity, let's recall what equicontinuity means.
Definition: Let $X$ and $Y$ be metric spaces. A set $E$ of continuous functions from $X$ to $Y$ (i.e. $E \subseteq C(X,Y)$ ) is equicontinuous if for every $x \in X$ and every $\varepsilon>0$ there exists $\delta >0$ such that
$$
y \in X \text{ and } d_X(x,y) < \delta \Rightarrow d_Y(f(x),f(y)) < \varepsilon \text{ for all } f \in E.
$$
With this definition in hand, we can state the result.
Theorem: Let $X$ and $Y$ be metric spaces with $X$ compact. Suppose that for $n \in \mathbb{N}$ the function $f_n: X \to Y$ is continuous and that $\{f_n | n \in \mathbb{N}\}$ is equicontinuous. Let $f : X \to Y$ and suppose that $f_n \to f$ pointwise as $n \to \infty$. Then $f_n \to f$ uniformly as $n \to \infty$. In particular, this means that $f$ is continuous.
Proof:
Let $\varepsilon >0$. By the equicontinuity property, for each $x \in X$ we can find $\delta_x >0$ such that
$$
y \in B_X(x,\delta_x) \Rightarrow d_Y(f_n(x),f_n(y)) < \frac{\varepsilon}{4} \text{ for all } n \in \mathbb{N}.
$$
The collection of open balls $\{B_X(x,\delta_x)\}_{x \in X}$ is an open cover of $X$. The compactness of $X$ allows us to find a finite subcover, namely $x_1,\dotsc,x_k \in X$ such that $X = \bigcup_{i=1}^k B_X(x_i,\delta_{x_i})$. Due to the pointwise convergence $f_n \to f$ (or really the pointwise Cauchy property), for each $i=1,\dotsc,k$ we can choose $N_i \in \mathbb{N}$ such that
$$
m,n \ge N_i \Rightarrow d_Y(f_n(x_i),f_m(x_i)) < \frac{\varepsilon}{4}.
$$
Set $N = \max \{N_1,\dotsc,N_m\} \in \mathbb{N}$. For each $x \in X$ there exists $i\in \{1,\dotsc,k\}$ such that $x \in B(x_i,\delta_{x_i})$. Then for $m \ge n \ge N$ we have the estimate
$$
d_Y(f_m(x),f_n(x)) \le d_Y(f_m(x),f_m(x_i)) + d_Y(f_m(x_i),f_n(x_i)) + d_Y(f_n(x_i),f_n(x)) \\
< \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{4} = \frac{3\varepsilon}{4}.
$$
Hence for $n \ge N$,
$$
d_Y(f(x),f_n(x)) = \lim_{m \to \infty} d_Y(f_m(x),f_n(x)) \le \frac{3\varepsilon}{4}.
$$
This holds for all $x \in X$, so
$$
n \ge N \Rightarrow \sup_{x \in X} d_Y(f(x),f_n(x)) \le \frac{3\varepsilon}{4} < \varepsilon,
$$
and we deduce that $f_n \to f$ uniformly as $n \to \infty$.
Best Answer
A significant difference was not mentioned in the comments: In the statement of Lemma 10, $A$ depends on $\eta$. Let us forget about closedness for a minute (I'll come back to that later) and compare the two statements:
Egoroff: $$(\forall \varepsilon>0) (\exists A\subseteq E), m(E\setminus A)<\varepsilon \boldsymbol{\large( \forall \eta>0)}$$ $$(\exists N\in\mathbb N)(\forall x\in A)(\forall n\geq N) |f_n(x)-f(x)|<\eta.$$
Lemma 10: $$(\forall \varepsilon>0) \boldsymbol{\large( \forall \eta>0)} (\exists A\subseteq E), m(E\setminus A)<\varepsilon $$ $$(\exists N\in\mathbb N)(\forall x\in A)(\forall n\geq N) |f_n(x)-f(x)|<\eta.$$
I know this isn't a pretty way to write it, but I hope it makes clear the important difference between the two. Notice that in Egoroff's theorem, $A$ had to be chosen once and for all to work for all (subsequently chosen) $\eta>0$, whereas in Lemma 10 $A$ only needs to work for a previously fixed $\eta$.
I remember when I was first learning measure theory from Royden doing an exercise showing how Egoroff's theorem can be proved using Lemma 10 (or whatever it was numbered in my edition). It was pretty straightforward because the exercise told me what sequences of $\delta$s and $\eta$s to use.
As for closedness, note that Egoroff's theorem without the assumption of closedness applies to all finite measure spaces (where there need not even be a topology in general) and the closedness can be added using inner regularity. For example, a subset $A$ of $\mathbb R$ is Lebesgue measurable if and only if for all $\varepsilon>0$ there exists a closed set $F\subseteq A$ such that $m^*(A\setminus F)<\varepsilon$ (where $m^*$ denotes Lebesgue outer measure). Uniform convergence on $A$ implies uniform convergence on F, and $E\setminus F = (E\setminus A)\cup (A\setminus F)$.