How many pairwise non-homeomorphic non-empty closed subsets of the Cantor set are there? My question is more or less related to basic set theory. But I don't know even that. Apologies if I added the wrong tags.

Motivation: How many non-compact (planar) surfaces are there upto homeomorphism?


Question: How many pairwise non-homeomorphic non-empty closed subsets of the Cantor set are there?

My idea is to produce an uncountable family $\mathcal F$ of closed subsets of the Cantor sets such that any two distinct elements of $\mathcal F$ are non-homeomorphic. Once I show this, the rest follows from the fact below:
Let $\mathcal P_1,\mathcal P_2$ be two non-empty, closed subsets of the Cantor set. Then $\Bbb S^2\setminus \mathcal P_1$ is homeomorphic to $\Bbb S^2\setminus \mathcal P_2$ if and only if $\mathcal P_1$ is homeomorphic to $\mathcal P_2$. This is a very particular case of Kerékjártó's classification theorem of non-compact surfaces.

More generally, I have the following question:
Let $\mathcal F'$ be the collection of all pairs $(\mathcal P,\mathcal A)$, where $\mathcal P$ is a non-empty closed subset of the Cantor set, and $\mathcal A$ is a closed subset of $\mathcal P$. Define an equivalence relation $\sim$ on $\mathcal F'$ as follows: $(\mathcal P_1, \mathcal A_1)\sim (\mathcal P_2, \mathcal A_2)$  if and only if there is a homeomorphism $\varphi\colon \mathcal P_1\to \mathcal P_2$ with $\varphi(\mathcal A_1)=\mathcal A_2$.

Question: What's the cardinality of $\mathcal F'/\sim$?

 A: There are $2^{\aleph_0}$ different subsets of the Cantor set up to homeomorphism.
There can't be more than $2^{\aleph_0}$ of them because any subset of the Cantor set is separable. To construct $2^{\aleph_0}$ of them, consider first the ordinal spaces $A_n=\omega^n+1$, with their order topologies. All of them can be imbedded in the Cantor set, and $A_n$ has Cantor Bendixon rank $n+1$.
Now, given any subset $S$ of $\mathbb{N}$, consider the usual, ternary Cantor set $C\subseteq[0,1]$ and the sequence $a_n=1-\frac{1}{3^n}$, so that the intervals $I_n=(a_n-\frac{1}{3^n},a_n]$ intersect $C$ only in $a_n$. For each $n\in S$ we can imbed a copy of $A_n$ in $I_n$, with $\omega^n$ identified to the point $a_n\in C$.
The resulting space $X_S\subseteq[0,1]$ is compact (because all ordinals are compact and the intervals $I_n$ converge to $1$). So as it is totally disconnected, it is homeomorphic to a closed subset of the Cantor set.
We just need to prove that $S$ can be obtained from the topology of $X_S$, so that different subsets of $\mathbb{N}$ give rise to different spaces up to homeomorphism. To that end, consider the set $Y_S\subseteq X_S$ formed by points which are not condensation points, and let $Z_S=\overline{Y_S}$ be its closure. $Z_S$ contains $Y_S$, the points $a_n$, with $n\in\mathbb{S}$, and no more points of $X_S$ (except for $1$ if $S$ is infinite). Moreover, the Cantor Bendixon rank of $a_n$ in $Z_S$ is $n+1$, and if $1$ is in $Z_S$ then its Cantor Bendixon rank is $\infty$, so yes, $S$ can be obtained from the Bendixon ranks of the points of $Z_S\setminus Y_S$.
