You can find encodings in the natural numbers of certain collections of subsets of $\mathbb{N}$. For example, as was pointed out by Henning Makholm, there is a very nice encoding of the collection of finite subsets of $\mathbb{N}$.
More generally, let $A$ be a recursively enumerable subset of $\mathbb{N}$. Let $T_A$ be a Turing machine that, on input $n$, halts if $n\in A$, and does not halt otherwise. Then we can encode $A$ by using the usual index of the machine $T_A$.
This encoding is not fully satisfactory. There are infinitely many Turing machines that halt on input $n$ iff $n\in A$. Thus every recursively enumerable set has infinitely many encodings. Moreover, there is no algorithm for telling, given two natural numbers, whether they encode the same set.
For more modest collections than the collection of recursively enumerable sets, there are far more satisfactory encodings. As a small and not very interesting example, consider the collection of subsets of $\mathbb{N}$ that are either finite or cofinite (their complement is finite). A small modification of the encoding of finite subsets will take care of the finites and cofinites. Basically, just use the even natural numbers for the finites. If you have used $2k$ to encode a finite, use $2k+1$ to encode its complement.
In addition, the elements of any countably infinite subset of the power set of $\mathbb{N}$ can, by the definition of countably infinite, be encoded using the natural numbers. However, by cardinality considerations, we cannot encode all the subsets of $\mathbb{N}$ using elements of $\mathbb{N}$.
You’ve misunderstood a couple of things. First, it’s not true that the Cantor-Bendixson derivatives of a countable set of reals necessarily vanish: every C-B derivative of $\Bbb Q$ is $\Bbb Q$, since $\Bbb Q$ has no isolated points to remove.
Secondly, a space is scattered if every subset contains at least one point that is isolated in that subset considered as a space in its own right. A simple sequence with its limit point is scattered: if the limit point is $p$, the point $p$ is isolated in the set $\{p\}$.
It’s true, however, that a closed subset of $\Bbb R$ is scattered (equivalently, has vanishing C-B derivative) if and only if it is countable.
First, no uncountable subset of $\Bbb R$ is scattered. This follows from the fact that $\Bbb R$ is second countable. Let $\mathscr{B}$ be a countable base for the topology, and let $A\subseteq\Bbb R$ be uncountable. Let $$\mathscr{B}_0=\{B\in\mathscr{B}:B\cap A\text{ is countable}\}\;,$$ and let $A_0=A\setminus\bigcup\mathscr{B}_0$. Clearly $\bigcup\mathscr{B}_0$ is countable, so $A_0$ is uncountable. If $x\in A_0$, and $U$ is any open nbhd of $x$, then there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U$. Clearly $B\notin\mathscr{B}_0$, so $B\cap A$ is uncountable, and therefore $B\cap A_0$ is uncountable as well. In particular, $x$ is not isolated in $A_0$. Thus, $A_0$ has no isolated points, and $A$ is not scattered.
Secondly, if $A\subseteq\Bbb R$ is countable and not scattered, then $A$ contains a countable infinite subset $A_0$ with no isolated points. Such a set is order-isomorphic to $\Bbb Q$ and therefore not closed.
Best Answer
As mentioned in the comments, the Cantor set has cardinality $2^{\aleph_0}$ (i.e. the same cardinality as the reals). Thus its power set has cardinality $2^{2^{\aleph_0}},$ which by Cantor's theorem is greater than the cardinality of the reals. On the other hand, it can be shown that there are only $2^{\aleph_0}$-many Borel sets (this isn't trivial). Thus, there are more subsets of the Cantor set than there are Borel sets, so there is a subset of the Cantor set that is not a Borel set.