Confusion on Derived Sets and the $n$-the Derived Set

I want to solve the following exercise 1.3.C. from page 27 of R. Engelking: General Topology.

For every positive integer $n$ the $n$-th derived set $A^{(n)}$ of a subset $A$ of a topological space $X$ is defined inductively by the formulas: $$ A^{(1)} = A^d \quad \textrm{and} \quad A^{(n)} = (A^{(n-1)})^d. $$ (a) Give an example of a set of real numbers that has three consecutive derived sets distinct from each other.

(b) Give an example of a set of real numbers that has infinitely many derived sets distinct from each other.

The definition is $$ A^d := \{ x \in X \mid x \in \overline{A \setminus \{ x \}} \}. $$ I am confused, cause the set $A^d$ ist closed, i.e. $\overline{A^d} = A^d$. And closed means it already contains all its limit points, so how can then $(A^d)^d$ be different from $A^d$?

To address your confusion about the closedness of derived sets: Note that $A^d$ being closed simply means that $(A^d)^d$ is a subset of $A^d$. (Actually $A^d\subseteq A$ is equivalent to $A$ being closed for any set $A$.) So this says that a limit point of $A^d$ is also a limit point of $A$, but a limit point of $A$ need not be a limit point of $A^d$, or in other words, $(A^d)^d$ can be a proper subset of $A^d$.

Now regarding your question about a subset with infinitely many distinct derived sets, try the following.

Let $B=\{0\}\cup\{\frac1n\mid n\in\mathbb N\}$.

For $n\ge2$ define $B_n^1=\{\frac1n\}\cup\{\frac1n+\frac1k(\frac1{n-1}-\frac1n)\mid k\ge2\}$ and let $b_n^1$ be the point $\frac1n+\frac12(\frac1{n-1}-\frac1n)$, the right most point of this sequence.

Define $B_n^{l+1}$ for $1\le l<n$ recursively as $B_n^{l+1}=\{b_n^l+\frac1k(\frac1{n-1}-b_n^l)\mid k\ge2\}$
and $b^{l+1}_n$ as the point $\frac12(\frac1{n-1}-b_n^l)$

Now you can take the union $$B\ \cup\ \bigcup\limits_{1 \leq l < n \\ n \in \mathbb{N}}{B_n^l}$$

If I'm not mistaken this should yield a set with infinitely many distinct derived sets.

You should draw a picture for the first few $l$ to get an idea how this set will look like.

A start: Consider first the set $P$ consisting of $0$ and all $\frac{1}{n}$, where $n$ ranges over the integers $\ge 2$.

The derived set of $P$ is $\{0\}$.

Now for every positive element $\frac{1}{n}$ of $P$, add a sequence that approaches $\frac{1}{n}$ from the right. For definiteness, the sequence for $\frac{1}{n}$ consists of the numbers $\frac{1}{n}+\frac{1}{k}\left(\frac{1}{n-1}-\frac{1}{n}\right)$, where $k$ ranges over the integers $\ge 2$.

Let $Q$ be $P$ together with these added points. The derived set of $Q$ is $P$, and the derived set of $P$ is $\{0\}$. This example should answer your $(A^d)^d$ question, and give a start towards the construction of examples that the problem asks for.

Note for example that $Q^d$, $(Q^d)^d$, and $((Q^d)^d)^d$ are distinct.