Let $f:[0,1]\to [0,1]$ be a continuous increasing function. Consider
$$x_*=\inf\{x\in [0,1]| f(x)\ge y\}.$$ If $x_*\not\in \{x\in [0,1]| f(x)\ge y\}$ then it is $f(x_*)<y.$ Now, by definition of infimum, for all $n\in\mathbb{N}$ there exists $x_n\in \{x\in [0,1]| f(x)\ge y\}$ such that $x_n\le x_*+1/n.$ So using that $f$ is increasing we get
$$y\le f(x_n)\le f\left(x_*+\frac 1n\right).$$ Since $f$ is continuous taking limits we get
$$y\le \lim_{n\to \infty}f\left(x_*+\frac 1n\right)=f(x_*).$$ Thus we get a contradiction. So it is
$$x_*=\inf\{x\in [0,1]| f(x)\ge y\}\in \{x\in [0,1]| f(x)\ge y\},$$ from where,
$$f(x_*)\ge y.$$
Edit
We will show that $f(x_*)=y$ under the additional assumption that $f(0)=0, f(1)=1.$
First of all, if $y=0$ then $x_*=0$ and we are done.
So we will suppose $0<y<1$ (and thus $x_*>0$). Assume that $f(x_*)>y.$ Since $f$ is continuous there exists $z\in (0,x_*)$ such that $f(z)=y.$ This contradicts the definition of $x_*.$ Thus, we have shown that $f(x_*)=y.$
Finally, consider $y=1.$ Since we have shown that $f(x_*)\ge y=1$ and $0\le f\le 1$ it is $f(x_*)=y=1.$
He's not noting that there are two copies of the Cantor set inside $[0,1] \cup [2,3]$; he's noting that the version of the Cantor set obtained by beginning from $[0,3]$ instead of $[0,1]$ is made up of two copies of the original Cantor set. Put more precisely:
The Cantor set is constructed by beginning with $C_0 = [0,1]$ and obtaining $C_{n + 1}$ by removing the middle third of each interval in $C_n$. Then the Cantor set is the set $C = \bigcap_iC_i$.
The tripled Cantor set is constructed by beginning with $\hat{C}_0 = [0,3]$ and obtaining $\hat{C}_{n + 1}$ by removing the middle third of each interval in $\hat{C}_n$. Then the tripled Cantor set is the set $\hat{C} = \bigcap_i\hat{C}_i$.
Now, $\hat{C}$ is clearly three times as "wide" as $C$ (it stretches across three times as much space). Because the process is recursive on the intervals, $\hat{C}\cap[0,1]$ is identical to $C$, and $\hat{C} \cap [2,3]$ is a translation of $C$. Therefore, $\hat{C}$ consists of two copies of $C$.
Note that the part that goes "$\hat{C}$ is clearly three times as 'wide' as $C$" is very loose - what does "wide" mean here? That's not an accident - this is a purposefully loose "definition" of dimension, because the actual, precise definition is difficult to understand without a warm-up.
Best Answer
The Cantor function is continuous and monotone increasing. It is constant on each middle ternary interval with a value which is a dyadic rational. Any inverse is therefore discontinuous at dyadic rationals and you have to decide what the value of the inverse should be on those points. Some possible choices: $$ \phi_-(y) = \sup \{x\in[0,1]: c(x)<y \} \leq \phi_+(y)= \inf \{x\in [0,1] : c(x) >y \} $$ which pick respectively the minimum and maximum possible inverse value at dyadic rationals.