In the Wikipedia of the Cantor function, one iterative construction for it is presented there:
Let $f_0(x)=x$ on $[0,1]$. For every positive integer $n$, define
$$f_{n+1}(x)=\begin{cases} \frac{f_n(3x)}{2}, & x\in[0,\frac{1}{3}]; \\ \frac{1}{2}, & x\in(\frac{1}{3},\frac{2}{3}); \\ \frac{1+f_n(3x-2)}{2}, & x\in[\frac{2}{3},1]. \end{cases}$$ Then $(f_n)$ converges uniformly to the Cantor function on $[0,1]$.
I somewhat noticed a interesting statement at the end of that section:
Also the choice of starting function does not really matter, provided $f_0(0)=0$, $f_0(1)=1$ and $f_0$ is bounded.
where a "citation needed" link is attached to the end of this statement. Therefore, I began to seek for any explanations, but most of the papers (such as this classical survey) only take this fact for granted without any further justification. I have come up with one solution, which is attached below, and it would be nice if this would attract more beautiful and concise solutions.
Best Answer
Endow the space $C[0,1]$ of continuous functions on the unit interval with the maximum norm.
Consider the transformation $T:C[0,1] \to C[0,1]$ defined by
$$(Tf)(x)=\begin{cases} \frac{f (3x)}{2}, & x\in[0,\frac{1}{3}]; \\ \frac{1}{2}, & x\in(\frac{1}{3},\frac{2}{3}); \\ \frac{1+f(3x-2)}{2}, & x\in[\frac{2}{3},1] \,, \end{cases}$$ so that $f_{n+1}=Tf_n$ for all $n$. then it is easy to verify that $$\forall f,g \in C[0,1], \quad \; \|Tf-Tg\| =\frac12 \|f-g\|\,.$$
By https://en.wikipedia.org/wiki/Banach_fixed-point_theorem , for any choice of $f_0$, the iterates $f_n=T^nf_0$ converge uniformly to the unique fixed point of $T$. Ths fixed point is the Cantor function $G$ due to the choice $f_0(x)=x$.
Edit: As noticed by the OP, the proof above applies when $f_0$ is continuous. If all we know is that $f_0$ is bounded, the same argument can be used in the space $B[0,1]$ of bounded functions with the supremum norm.