Characterization of real-valued $C^1$ functions on $[0,1]$ I have encountered the following example, where I found things I don't understand:

Let $\Bbb{Q}^+$ be the set of positive rational numbers, $C([0,1])$ be the space of all continuous real-valued functions on $[0,1]$ and denote by $C^1$ the class of continuously differentiable functions in $C([0,1])$.
  (At the endpoints we take one-sided derivatives.)
  Then for $f\in C([0,1])$, $f\in C^1$ iff for all $\epsilon\in\Bbb{Q}^+$ there exists rational open intervals $I_0,\dots,I_{n-1}$ covering $[0,1]$ s.t. for all $j<n$:
  $$\forall a,b,c,d\in I_j\cap [0,1], a\ne b,c\ne d,\lvert \frac{f(a)-f(b)}{a-b}-\frac{f(c)-f(d)}{c-d}\rvert\le \epsilon.$$
  So for an open interval $J$ and $\epsilon>0$, we put
  $$A_{J,\epsilon}=\{f\in C([0,1])\mid \forall a,b,c,d\in J\cap [0,1], a\ne b,c\ne d,\lvert \frac{f(a)-f(b)}{a-b}-\frac{f(c)-f(d)}{c-d}\rvert\le \epsilon\},$$
  we have that $A_{J,\epsilon}$ is closed in $C([0,1])$.

As I understand it, functions in $C^1$ are characterized in terms of the uniform continuity of the quotient difference (right?).
Question: How can I prove that?
Attempt: By definition, $$f\in C^1 \iff \forall x\in (0,1),\exists \lim_{y\to x}\frac{f(y)-f(x)}{y-x}=f'(x), f'(x)=\lim_{y\to x}f'(y)$$
(at endpoints a similar condition holds with right/left limits).
It seems clear to me that the above implies continuity of the quotient difference at every $x\in [0,1]$, but does the converse hold?
If not, how can I rigorously prove the above characterization?
Thank you in advance for your help.
 A: 
It seems clear to me that the above implies continuity of the quotient difference at every $x\in [0,1]$, but does the converse hold?

Indeed, the clear implication follows from Lagrange’s theorem and compactness of $[0,1]$. To show the converse impliction suupose that the function $f\in C[0,1]$ satisfies the given interval condition. Let’s prove that $f\in C^1[0,1]$. Let $x\in [0,1]$ be an any point. For any $\varepsilon>0$ there exists a rational open interval (as a rational open interval I shall understood an open in $[0,1]$ interval with rational endpoints) $J(\varepsilon)$ such that $f\in A_{ J(\varepsilon),\varepsilon}$. In particular, for each $b,d\in J(\varepsilon)\setminus\{x\}$ we have $\lvert \frac{f(x)-f(b)}{x-b}-\frac{f(x)-f(d)}{x-d}\rvert\le \varepsilon$. This easily implies that there exists a limit $\lim_{b\to x} \frac{f(x)-f(b)}{x-b}$, that is $f’(x)$. Moreover, $\lvert \frac{f(x)-f(b)}{x-b}-f’(x) \rvert\le \varepsilon$ for each $b\in J(\varepsilon)\setminus\{x\}$. Similarly, if $y\in J(\varepsilon)$ then $\lvert \frac{f(y)-f(d)}{y-d}-f’(y) \rvert\le \varepsilon$ for each $d\in J(\varepsilon)\setminus\{y\}$. This easily implies that $\lvert f’(x)-f’(y) \rvert\le 3\varepsilon$, so the function $f’$ is continuous on $[0,1]$.
