I think that your proof is not correct, because you are using that the derivative is Riemann integrable. If $f'$ were continuous, then you could apply the fundamental theorem of calculus (like you did) and the result would follow.
But there are examples of differentiable functions with bounded derivative, such that their derivative is not Riemann integrable. One of these examples is Volterra's function.
I'll prove what you are asking, the way you are asking, but only for $c \in \mathbb Q$. For irrationals, a similar argument follows.
You want to find an $\epsilon>0$ such that $\forall \delta > 0$, there is an $x \in I$, such that $|x-c| < \delta$ but $|f(x)-f(c)| > \epsilon$.
So we will take $\epsilon = \frac 12$.
For any $\delta > 0$, pick an irrational number $x \in (c, c - \delta)$ (we can do this however small $\delta$ is, as long as it is positive). Then, since $x \notin \mathbb Q, f(x) = 0$, so $|f(x)-f(c)| = 1 > \epsilon$ although $|x - c| < \delta$. This shows discontinuity at $c \in \mathbb Q$.
The is called the epsilon-delta definition of continuity, while the definition being used there is called the sequential definition of continuity. You can show these are equivalent, but first read them up.
The sequential definition is the following: $f$ is continuous if whenever $x_n$ is a sequence, $x_n \to c$, then $f(x_n) \to f(c)$. ($\to$ means "converges to").
So in this case, you can find an sequence of irrationals $c_n$ converging to $c$, however $f(c_n) = 0$ in that case, but $f(c) = 1$, and clearly the constant sequence $0$ doesn't converge to $1$, which is a contradiction. This is what the author is trying to say.
Hope you have understood. Do reply back.
Best Answer
Hints: assuming the OP means $[a,b]$, (because the claim is false as stated (see comment below):
Let $x_0\in [a,b].$
If $\epsilon>0,$ then there is a $\delta>0$ such that $|x-x_0|<\delta\Rightarrow |f(x)-f(x_0)|<\frac{\epsilon}{2}.$ It follows that if $x_0<x,\ $ and $|x-x_0|<\delta,$
and therefore, using the definition of the supremum,
Similarly, if $x<x_0$ and $|x-x_0|<\delta,$ then $f(x_0)<f(x)+\frac{\epsilon}{2}$ and so