As Jose27 noted, uniformly continuous functions need not be differentiable even at a single point.
It is true that if $f$ is defined on an interval in $\mathbb R$ and is everywhere differentiable with bounded derivative, then $f$ is uniformly continuous. In fact, it follows from the Mean Value Theorem that such an $f$ is Lipschitz, which is much stronger.
However, if $f$ is uniformly continuous and everywhere differentiable, then $f$ need not have bounded derivative. Jose27 mentions $\sqrt x$, which would work as an example on the interval $(0,\infty)$. The function
$$f\left(x\right)=\begin{cases}x^2\sin\left(\frac{1}{x^2}\right) & :x\neq 0\\ 0 &:x=0\end{cases}$$
is uniformly continuous on any bounded interval such as $(-1,1)$, but has unbounded derivative near $0$. There are also examples where $f'$ is bounded on bounded intervals, but unbounded on $\mathbb R$, while $f$ is uniformly continuous. You can show that any continuous function $f$ on $\mathbb R$ such that $\lim\limits_{|x|\to \infty}f(x)=0$ is uniformly continuous, and using this fact you can see that Nate Eldredge's example here of $\sin(x^4)/(1+x^2)$ provides such an example. Another source of examples is the question
Why if $f'$ is unbounded, then $f$ isn't uniformly continuous?
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
Hint: Is it not the case that $f:\mathbb{R}\to \mathbb{Z}$ given by $f(x)=1$ and $g:\mathbb{R}\to \mathbb{Z}$ given by $g(x)=0$ are both continuous?