It's called the middle-thirds Cantor set because in general you can construct a class of sets with similar properties using a similar but scaled construction. For example, you can start with $[0,1]$, remove an interval of length $\frac{1}{2}$ from the center. Then you have $2$ intervals of length $\frac{1}{4}$, remove an interval of length $\frac{1}{8}$ from the center. Inductively, at each step you have a disjoint union of intervals of length $l$ remaining in your set, and you remove an interval of length $l/2$ from each interval. This set, which we might call the Cantor "middle-halves" set, has many of the same properties of the Cantor middle-thirds set. From this you can imagine constructing the "middle-fourths" set and many other Cantor-type sets. The middle-thirds set is sort of standard because it's the easiest Cantor-type set to construct (mainly it's easy to figure out the lengths of the intervals at each step).
The reason for the ternary expansion is precisely as you stated, the middle-thirds construction removes any numbers with absolutely necessary $1$-s in the ternary expansion. You should check this for yourself for a few cases: for example, check if $0.1abcd...$ (ternary) can be in the middle-thirds set, then $0.01abcd...$, $0.21abcd...$, and so on. Then you'll see why the Cantor set construction removes them. This excludes cases like $1/3$, which lies in the Cantor set and has ternary expansion $0.1000...$, because it can also be written $0.0222...$ .
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
Yes, it is. You can see this very neatly with an explicit map. Let $K$ denote the Cantor set. Given an element $$x=\sum_{n=1}^\infty\frac{2a_n}{3^n}\in K$$ where $a_n=0$ or $1$ for each $n$, let $$f(x)=\sum_{n=1}^\infty\frac{a_n}{2^n}.$$ That is, $f$ takes the ternary expansion of $x$ using $0$s and $2$s, replaces each $2$ with a $1$, and considers it as a binary expansion. Then $f:K\to [0,1]$ is a surjection that is easily checked to be continuous. Since $K$ is compact and $[0,1]$ is Hausdorff, it follows that $f$ is a quotient map. Finally, the corresponding equivalence relation is exactly the one you describe, since the pairs you are identifying are exactly the pairs whose ternary expansions correspond to the two different binary expansions of some dyadic rational.