The Cantor set is homeomorphic to infinite product of $\{0,1\}$ with itself - cylinder basis - and it topology

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 $b = (b_n)_n \in X$, $$h^{-1}(b) = \sum_{n=1}^\infty\frac{2b_n}{3^n}.$$

Note that the $1/3$-Cantor set in $[0,1]$ can be represented as the set of real numbers of the form $\sum_{n=1}^\infty a_n/3^n$ where $a_n\in\{0,2\}$ for each $n\in\mathbb{N}$. A homeomorphism you are looking for is the function $f$ which maps the point $\sum_{n=1}^\infty a_n/3^n$ in the Cantor set to the sequence $(a_n/2)_{n=1}^\infty$ in the product $\{0,1\}^\mathbb{N}$. The product $\{0,1\}^\mathbb{N}$ consists of countably infinite sequences of $0$'s and $1$'s. Note that no finite tuple such as $(0,0,0,1)$ is in $\{0,1\}^\mathbb{N}$. The product is topologized so that each factor $\{0,1\}$ is given the discrete topology and then the product is given the product topology.

You want to prove that $f$ is a continuous and open bijection. The bijectiveness is very easy to show. For the continuity you may want to use the fact that the product topology of $\{0,1\}^\mathbb{N}$ is generated by the sets of the form $U(N,a)=\{(a_n)_{n=1}^\infty\in\{0,1\}^\mathbb{N}:a_N=a\}$ where $N\in\mathbb{N}$ and $a\in\{0,1\}$, and hence it suffices to show that the preimages of these sets $U(N,a)$ are open in the Cantor set. Finally to show that $f$ is open you can use the following general fact: a continuous bijection from a compact space to a Hausdorff space is open.

The Cantor set consists of numbers whose ternary expansion uses only $0$s and $2$s. So there's a "natural" bijection between the cantor set and $\{0,1\}^\omega$, or rather $\{0,2\}^\omega$. Everything else should just "work out".

Note that $\{0,1\}^\omega$ consists of all infinite sequences of $0$ and $1$.