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 $b = (b_n)_n \in X$, $$h^{-1}(b) = \sum_{n=1}^\infty\frac{2b_n}{3^n}.$$
Your space $C$ is the collection of functions $f:\omega=\{0,1,2,\ldots\}\to \{0,1\}=X$, i.e. the collection of infinite binary sequences. I will try to illustrate how we can understand the space $C$ and its topology below, deriving some of its properties.
The basic open sets in this collection are (by definition) infinite products of open sets in $X$ where all but finitely many factors are $X$. Suppose the factors correspond to a sequence $k_1<k_2<\cdots<k_s$ of natural numbers. At these factors (assuming the set is not empty) there correspond either the open set $\{0\}$ or the open set $\{1\}$. So we can associate to each basic open set $O$ a function from a finite subset $F=\{k_1,\ldots,k_s\}$ of $\omega$ to $X$. This is a bijection. Note that if $f,g\in O$ and $F$ is associated to $O$; then $f\mid F=g\mid F$.
Now take a function $f:\omega\to X$. We can consider the open sets $O_i=\prod_{k=0}^i \{f(k)\}\times X\times \cdots$. Note that given any basic open set $O$ that contains $f$, there is $j$ such that $O_j\subseteq O$ (can you prove this?). This means that the collection $\mathscr B=\{O_i:i\in\omega\}$ is a neighborhood basis for $f$. This means our space is second countable! It is also separable, since the collection of functions that are eventually constant is countable and dense: given a basic neighborhood $O_i$ of $f$, the function that agrees with $f$ on $\{0,1,2,\ldots,i\}$ and is constant afterwards is in $O_i$. This proves this space is also perfect.
In particular $g\in \overline{\{f\}}$ iff $g$ agrees with $f$ on $\{0,1,\ldots,i\}$ with each $i$. This means $g=f$; so singletons are closed $\{f\}$. In fact your space is Hausdorff: if $g$ and $f$ do not agree at position $k$ (say one takes the value $0$ and the other $1$), then $\cdots\times X\times \{0\}\times X\times\cdots$ and $\cdots\times X\times \{1\}\times X\times\cdots$ are disjoint open sets that contain them.
Actually, your space is metrizable, and the neigborhood basis we obtained gives a decent idea of what this metric is: we define $d(f,g)=2^{-\nu(f,g)}$ where $\nu(f,g)$ is the first $k\in\omega$ such that $f(k)\neq g(k)$. We define $\nu(f,f)=\infty$, of course! Note then that $O_i=\{g\in C:d(f,g)<2^{-i}\}$. Check that $d(f,g)$ is indeed a metric.
Finally, your space is compact. Since we've shown it is metrizable, we can simply check it is sequentially compact. Now take a sequence of functions $(f_i)$ in your space. If we look at $f_i(0)$ for $i=1,2,3,\ldots$, we get a sequence of zeros and ones. Since this sequence is infinite, either the ones or the zeros repeat infinitely often. So we can take a subsequence indexed by $N_0$ that has all its first coordinates equal to say $x_0\in \{0,1\}$. Now look at the second coordinate of this sequence, and repeat the process to get $N_1\subseteq N_0$ and some $x_1$. We get a sequence of nested infinite subsets $N_0\supseteq N_1\supseteq N_2\supseteq \cdots$, and we may thus take a subsequence $f_{k_j}$ such that $k_i\in N_i-N_{i+1}$. You can check that this subsequence converges to the $f$ such that $f(i)=x_i$ (note this is just another diagonalization trick!).
Given we have exhibited your space as a metric space, I believe you can understand its topology quite nicely. Now, note that the Cantor set consists of real numbers whose ternary expansion contains only $0$s and $2$s. Does this give you an idea of what a possible homeomorphism $\eta:C'\to C$ may be?
Best Answer
To be explicit, here is the function I think you are referring to. Let $K$ denote the Cantor set. An element $x\in K$ has a base $3$ expansion consisting of $0$s and $2$s. Define $f(x)$ as the number whose binary expansion is obtained from the base $3$ expansion of $x$ by replacing the $2$s with $1$s.
So, why is this function $f:K\to [0,1]$ not injective? After all, you can "invert" $f$ by taking a binary expansion, replacing the $1$s with $2$s, and reading it as a ternary expansion. The answer is that binary expansions are not always unique. For instance, $1/2$ has both $$0.011111\dots$$ and $$0.100000\dots$$ as binary expansions. But when you "invert" $f$ on these, you get base $3$ expansions that do not represent the same number: $$0.022222\dots$$ is $1/3$ and $$0.200000\dots$$ is $2/3$. So we have $f(1/3)=f(2/3)=1/2$, and $f$ is not injective.
More generally, if you view $K$ as the interval $[0,1]$ with infinitely many "holes" cut out of it (the middle third intervals you remove), the function $f$ glues together the endpoints of each hole. Intuitively, after doing this, there are no holes left, so you get a genuine connected interval which is homeomorphic to $[0,1]$. (Of course, it is not obvious that this is actually true, and it takes some nontrivial work to prove it.)