This question regards Advanced Calculus of Several Variables, by C.H. Edwards, Jr.
Theorem VI.1.4 Let $\left\{ f_{n}\right\} _{1}^{\infty}$ be a sequence of continuously differentiable real-valued functions on $\left[a,b\right],$ converging (pointwise) to $f$. Suppose that the $\left\{ f_{n}^{\prime}\right\} _{1}^{\infty}$ converges uniformily to $g$. Then $\left\{ f_{n}\right\} _{1}^{\infty}$ converges uniformly to $f,$ and $f$ is differentiable, with $f^{\prime}=g$.
PROOF By the fundamental theorem of calculus, we have
$$ f_{n}\left[x\right]=f_{n}\left[a\right]+\int_{a}^{x}f_{n}^{\prime} $$ for each $n$ and each $x\in\left[a,b\right]$. From this and Exercise IV.3.4 (on the termwise integration of a uniformly convergent sequence of continuous functions) we obtain
$$ f\left[x\right]=\lim_{n\to\infty}f_{n}\left[x\right] $$
$$ =\lim_{n\to\infty}f_{n}\left[a\right]+\lim_{n\to\infty}\int_{a}^{x}f_{n}^{\prime}, $$
$$ f\left[x\right]=f\left[a\right]+\int_{a}^{x}g. $$ Another application of the fundamental theorem yields $f^{\prime}=g$ as desired.
To see that the convergence of $\left\{ f_{n}\right\} _{1}^{\infty}$ to $f$ is uniform, note that
$$ \left|f_{n}\left[x\right]-f\left[x\right]\right|=\left|\int_{a}^{x}f_{n}^{\prime}-\int_{a}^{x}g\right|+\left|f_{n}\left[a\right]-f\left[a\right]\right| $$
$$ \le\int_{a}^{x}\left|f_{n}^{\prime}-g\right|+\left|f_{n}\left[a\right]-f\left[a\right]\right| $$
$$ \le\left(b-a\right)\left\Vert f_{n}^{\prime}-g\right\Vert _{0}+\left|f_{n}\left[a\right]-f\left[a\right]\right|. $$ The uniform convergence of the sequence $\left\{ f_{n}\right\} _{1}^{\infty}$ therefore follows from that of the sequence $\left\{ f_{n}^{\prime}\right\} _{1}^{\infty}$.
I do not see how the equivalence
$$ \left|f_{n}\left[x\right]-f\left[x\right]\right|=\left|\int_{a}^{x}f_{n}^{\prime}-\int_{a}^{x}g\right|+\left|f_{n}\left[a\right]-f\left[a\right]\right| $$ is justified. The equation $$ f_{n}\left[x\right]-f\left[x\right]=\int_{a}^{x}f_{n}^{\prime}+f_{n}\left[a\right]-\left(\int_{a}^{x}g+f\left[a\right]\right) $$
$$ =\int_{a}^{x}f_{n}^{\prime}-\int_{a}^{x}g+f_{n}\left[a\right]-f\left[a\right] $$ is clearly valid. But the triangle inequality gives $$ \left|\left(\int_{a}^{x}f_{n}^{\prime}-\int_{a}^{x}g\right)+\left(f_{n}\left[a\right]-f\left[a\right]\right)\right|\le\left|\int_{a}^{x}f_{n}^{\prime}-\int_{a}^{x}g\right|+\left|f_{n}\left[a\right]-f\left[a\right]\right|. $$
I can think of no reason why the functions $f_n$, $f$ and their derivatives cannot, in general, change signs, or have their differences change signs on the interval $\left[a,b\right]$.
Is Edwards's statement of the proof correct?
It's a typo. As you say, it ought to be "$\leqslant$".