I am trying to prove the question in the title and believe that I have arrived at a $\frac{\epsilon}{3}$ argument. Could anyone take a look at it and provide any comments or insight, it seems an important theorem.
Suppose that $x \in [a,b]$ and $\epsilon > 0$, then since $f_n(x)$ converge uniformly to $f$, $\exists N$ s.t. $\forall n\geq N$
$|f_n(x) - f(x)| < \frac{\epsilon}{3}$, $\forall x \in [a, b], \forall n \geq N$. Also, since $f_N(x)$ is continuous $\exists \delta > 0$ so that if $y \in [a,b]$ and $|x-y| < \delta$
then $|f_N(y) - f_N(x)| < \frac{\epsilon}{3}$. Then we have for $y \in [a,b]$ and $|x-y| < \delta$ that
$ |f(x) - f(y)| = |f(x) - f_N(x) - f(y) + f_N(y) + f_N(x) - f_N(y)|$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \leq |f_N(x) -f(x)| + |f_N(y) - f(y)| + |f_N(x) + f_N(y)|$
$\ \ \ \ \ \ \ \ \ \ \ \ \ < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon $