I’m going to assume that Cantor set here refers to the standard middle-thirds Cantor set $C$ described here. It can be described as the set of real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$, i.e., real numbers of the form $$\sum_{n=1}^\infty \frac{a_n}{3^n},$$ where each $a_n$ is either $0$ or $2$.
For each positive integer $n$ let $D_n = \{0,1\}$ with the discrete topology, and let $$X = \prod_{n=1}^\infty D_n$$ with the product topology. Elements of $X$ are infinite sequences of $0$’s and $1$’s, so $(0,0,0,1)$ and $(0,1,1,1,1,1,1)$ are not elements of $X$; if you pad these with an infinite string of $0$’s to get $(0,0,0,1,0,0,0,0,\dots)$ and $(0,1,1,1,1,1,1,0,0,0,0,\dots)$, however, you do get points of $X$. A more interesting point of $X$ is the sequence $(p_n)_n$, where $p_n = 1$ if $n$ is prime, and $p_n = 0$ if $n$ is not prime.
Your problem is to show that $C$, with the topology that it inherits from $\mathbb{R}$, is homeomorphic to $X$. To do that, you must find a bijection $h:C\to D$ such that both $h$ and $h^{-1}$ are continuous. The suggestion that you found is to let $$h\left(\sum_{n=1}^\infty\frac{a_n}{3^n}\right) = \left(\frac{a_1}2,\frac{a_2}2,\frac{a_3}2,\dots\right).$$ Note that $$\frac{a_n}2 = \begin{cases}0,&\text{if }a_n=0\\1,&\text{if }a_n=2,\end{cases}$$ so this really does define a point in $X$. This really is a bijection: if $b = (b_n)_n \in X$, $$h^{-1}(b) = \sum_{n=1}^\infty\frac{2b_n}{3^n}.$$
Best Answer
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 $f(x_*)=y=1.$