In the first part of the proof, the author says $\lim_{x \to c} \frac {f(x)-f(c)}{x-c} = f^\prime(c) \ge 0$, but what we really want to get is $f^\prime (x) \ge 0$
So, is it okay if I say $\lim_{y \to x} \frac {f(y)-f(x)}{y-x} = f^\prime(x) \ge 0$?
The atuhor uses the notation, $c$, as a cluster point before, so I am not sure if it is allowed to replace c by x.
By definition of the derivative, $$f^\prime(c) = \lim_{x \to c} \frac {f(x)-f(c)}{x-c}.$$ The proof's argument is that $$\frac {f(x)-f(c)}{x-c}\ge 0,$$ for all $x$ and $c$ in $I$. Therefore, in particular, this is true in the limit as $x$ goes to $c$. Since that is the definition of $f'(c)$ to begin with, it follows that $$f'(c) \ge 0, $$ for all $c$ in $I$. The variable $c$ is just a dummy variable, if you prefer you can call it $y$.