Uniform convergence of derivatives, Tao 14.2.7.

Since $\{f_n(x_0)\}$ converges, for each $\epsilon > 0$ and $n, m$ large enough we have $$ \begin{align} \lvert f_n(x) - f_m(x) \rvert &\leq \left\lvert (f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0)) \right\rvert + \left\lvert f_n(x_0) - f_m(x_0) \right\rvert \\ &\leq \epsilon \left\lvert x - x_0 \right\rvert + \epsilon \\ &\leq \epsilon (b - a) + \epsilon \end{align} $$ Hence $f_n$ converges uniformly on $I$ to a function $f$, moreover for each $\epsilon > 0$ and $m, n$ large enough, the inequality $$ \left\lvert \frac {f_n(y) - f_n(x)} {y - x} - \frac {f_m(y) - f_m(x)} {y - x} \right\rvert \leq \epsilon $$ holds for each $x\neq y\in I$. (It is the same inequality of the hint but now we can assume it holds for generic $y\in I$, because we showed $f_n(y)$ converges for all $y \in I$)
The above relation implies that $\frac {f_n(y) - f_n(x)} {y - x}$ converges uniformly to $\frac {f(y) - f(x)} {y - x}$.

Now we can write $$ \left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq \\ \left\lvert\frac {f(y) - f(x)} {y - x} - \frac {f_n(y) - f_n(x)} {y - x} \right\rvert + \left\lvert \frac {f_n(y) - f_n(x)} {y - x} - f_n'(x)\right\rvert + \left\lvert f_n'(x) - g(x) \right\rvert $$ For each $\epsilon > 0$ and $n$ large enough we get $$ \left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq 2\frac \epsilon 3 + \left\lvert \frac {f_n(y) - f_n(x)} {y - x} - f_n'(x)\right\rvert $$ and for $y$ close enough to $x$ $$ \left\lvert\frac {f(y) - f(x)} {y - x} - g(x) \right\rvert \leq \epsilon $$ So $f'(x)$ exists and is equal to $g(x)$.

Edit

To clarify the point raised by @DavidC.Ullrich.

Since ${f'_n}$ converges uniformly, there exists $N \in \mathbb N$ such that $\lVert f'_n - f'_m \rVert_\infty < \epsilon$ for all $n, m > N$, that is $$ |f'_n(x) - f'_m(x)| < \epsilon \qquad \forall m,n > N, \forall x\in I $$

So, by means of the mean value theorem, for each $m,n > N$ and for each $x \neq y\in I$ we can write

$$ \left\lvert \frac {f_n(y) - f_n(x)} {y - x} - \frac {f_m(y) - f_m(x)} {y - x} \right\rvert = \\ \left\lvert \frac {f_n(y) - f_m(y)} {y - x} - \frac {f_n(x) - f_m(x)} {y - x} \right\rvert = \\ \left\lvert \frac {(f_n - f_m)(y)- (f_n - f_m)(x)} {y - x}\right\rvert = \\ \lvert (f_n - f_m)'(\xi) \rvert = \\ \lvert f_n'(\xi) - f_m'(\xi)\rvert < \epsilon $$

Due to the uniform convergence of the $f'_n$ you can find an $N$ for every $\epsilon$ such that $(f'_n(x) - g(x)) \leq \epsilon$ for all $n \geq N$, which is equivalent to $d_\infty(f'_n,g) \leq \epsilon$. Thus, $d_\infty(f_n',g) \to0 $ as $n\to\infty$.

Now, $\int_{a}^x f_n'(y) - g(y) dy \leq d_\infty(f_n',g)|x-a| \leq d_\infty(f_n',g)|b-a|$. Since $d_\infty(f_n',g) \to0 $ as $n\to\infty$ you get that $\int_a^x f_n'(y)dy$ converges uniformly to $\int_a^x g(y)dy$.

In general, that won't transfer to $f_n(x) = c_n + \int_a^x f_n'(y)dy$ because the $c_n$ could be chosen maliociously. But if $\lim_{n\to\infty}f_n(x) = \lim_{n\to\infty} c_n + \int_a^x f_n'(y)dy$ converges for one $x$, then $\lim_{n\to\infty}c_n$ must converge, since the second term converges too (every uniformly!).

Which in turn means the limit must actually converge for all $x$, ecause $\lim_{n\to\infty}c_n$ doesn't actually depend on $x$. And for the same reason (and because the other term converges uniformly), the convergence is even uniform.