Let
$F_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$
$F_2 = [0,\frac{1}{3^2}] \cup [\frac{2}{3^2},\frac{3}{3^2}] \cup [\frac{6}{3^2},\frac{7}{3^2}] \cup [\frac{8}{3^2},1]$
and so on.
The Cantor set is then $C=\bigcap^{\infty}_{k=1} F_k$.
Each $x \in C$ can be written in the form $ x = \frac{a_0}{3} + \frac{a_1}{3^2} + \frac{a_2}{3^3} + ...+ \frac{a_n}{3^{n+1}}+ ...$
By Construction, if $x\in F_k$ and $0 \leq j < k$, then $3^jx=a+y$ where $a \in \mathbb{N} \cup \{0\}$ (*) and $y \in F_{k-j}$.
Noting this pattern suppose $x\in C$ therefore $x \in F_k$ for all $k \in \mathbb{N}$. Now assume that for some $m\in \mathbb{N}$ we have $a_m = 1$ Then we have $3^mx=a+\frac{1}{3}+\frac{a_{m+1}}{3^2}+\frac{a_{m+2}}{3^3}+...\ \ \ \Leftrightarrow$ $ \ \ \ \ y=3^mx-a \notin F_1 \Leftrightarrow x \notin F_{m+1} \Leftrightarrow x \notin C$ contradiction.
To see the pattern more clearly, consider $F_3$,
$F_3= [0,\frac{1}{27}] \cup [\frac{2}{27},\frac{3}{27}] \cup [\frac{6}{27},\frac{7}{27}]\cup [\frac{8}{27},\frac{9}{27}]\cup [\frac{18}{27},\frac{19}{27}]\cup [\frac{20}{27},\frac{21}{27}] \cup [\frac{24}{27},\frac{25}{27}] \cup [\frac{26}{27},1]$
Define the "multiplication" of an interval by an scalar, naturally by multiplying the endpoints by that scalar, for example $3F_2= [0,\frac{3}{9}] \cup [\frac{6}{9},1] \cup [\frac{18}{9},\frac{21}{9}] \cup [\frac{24}{9},3]=[0,\frac{1}{3}] \cup [\frac{2}{3},1] \cup [2,\frac{7}{3}] \cup [\frac{8}{3},3]$
This multiplication is important, since it gives us 2 copies of $F_1$, (the 2nd copy is a translation of $F_1$ by an integer, look at (*) )
More generally, $3^jF_k$ gives us $2^j$ copies of $F_{k-j}$
let $x \in F_3$
For $j=0 \ \ \ $, the statement is clear, i.e. $3^0x \in F_{3-0} \Rightarrow x \in F_3$ Trivial !
For $j=1 \ \ \ $Obviously $3^1x \in 3F_3$, i.e. $3^1x$ lies in a translated (by a number $a \in \mathbb{N} \cup \{0\}$) copy of $F_{3-1}=F_2$ Therefore $3x=a+y$ where a is the natural number of translation, and $y\in F_2$
For $j=2 \ \ \ $ Similarly $3^2x \in 3^2F_3 \Rightarrow 3^2x $ lies in a translated copy of $F_{3-2}=F_1$
Consider a right-hand endpoint of one of the intervals removed to form the Cantor set. It has a ternary representation
$$x = 0.(2a_1)(2a_2)\ldots(2a_n)2000\ldots$$
where the $a$'s are all $0$ or $1$,
and the binary representation of $f(x)$ is
$$f(x) = 0.(a_1)(a_2)\ldots(a_n)1000\ldots.$$
Pick $m > n$ and $h>0$ with $3^{-(m+1)} < h < 3^{-m}.$
Then as $m \rightarrow \infty$ and $h \rightarrow 0+$
$$\frac{f(x+h)-f(x)}{h}>\frac{3^m}{2^{m+1}}\rightarrow \infty$$
and the right-hand derivative $f'_+(x) = \infty$.
You can make a similar argument for a left-hand endpoint of a removed interval.
Best Answer
Consider the metric space of continuous non decreasing functions $f$ on $[0,1]\to\Bbb R$ so that $f(0)=0$, $f(1)=1$ with the maximum norm. Then this space is complete (as the space of continuous functions is complete and by pointwise limit the conditions on $f(0)$ and $f(1),$ as well as inequalities, carry on to limits).
The map $\Gamma$ where $$\Gamma(f)(x)=\begin{cases}\frac12 f(3x),&\text{ if }x\in\left[0,\frac13\right)\\\frac12,&\text{ if } x\in\left[\frac13,\frac23\right]\\\frac12(1+f(3x-2)),&\text{ if }x\in\left(\frac23,1\right]\end{cases}$$ (it is easy to see that this satisfies the above conditions) is a strict contraction, as $$ \|\Gamma f-\Gamma g\|_\infty \le \frac12\|f-g\|_\infty$$
Thus, by the Banach fixed point theorem, the function $c$ from the second definition is in fact the unique fixed point of $\Gamma$. Thus, we get $$ \Gamma(c) = c $$
So, if $a_0$ is the first ternary digit of $x$, then this shows us: $$ c(x) = \frac12\left(\frac{a_0}2+c(3x-a_0)\right) $$ if $a_0\neq 1$ or $c(x)\frac12$ if so. By iteration on $c(3x-a_i)$ this shows us that if $I$ is the first time that the terniary digit $a_I$ of $x$ is $1$ we have $$ c(x) = \sum_{i<I}\frac{a_i}{2^{i+1}} + \frac1{2^I}$$ and if this never happens (i.e. $x\in C$) we get by taking limits $$ c(x) = \sum_{i\in\Bbb N} \frac{a_i}{2^{i+1}}$$
Note that as $c$ is continous and non-decreasing we get if $x\not \in C$ that $\sup\limits_{y<x,y\in C} c(y) = c(x)$.
Thus we have proven that $c$ has the form of the first definition.