Here's Prob. 13 in the Exercises after Chap. 2 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition.
Construct a compact set of real numbers whose limit points form a countable set.
Here's Def. 2.4(c) in Rudin.
For any positive integer $n$, let $J_n$ be the set whose elements are the integers $1, 2, \ldots, n$; let $J$ be the set consisting of all positive integers. For any set $A$, we say $A$ is countable if $A \sim J$ (i.e. there is a $1$-$!$ correspondence between $A$ and $J$).
I know that this question has been asked many times here before. However, I would like to demonstrate my effort, which goes as follows.
For each positive integer $n$, let the set $A_n$ be defined as follows. $$A_n \colon= \left\{ \ \frac 1 n - \frac 1 k \ \colon \ k \in \mathbb{N}, \ k > n(n+1) \ \right\}.$$ Thus, we have $$A_1 = \left\{ \ \frac 2 3, \frac 3 4, \frac 4 5, \frac 5 6, \ldots \ \right\},$$ $$A_2= \left\{ \ \frac{5}{14}, \frac{3}{8}, \frac{7}{18}, \frac{2}{5}, \frac{9}{22}, \frac{5}{12}, \frac{11}{26}, \frac{3}{7}, \frac{13}{30}, \frac{7}{16}, \ldots \ \right\},$$ and so on
Now let $$A_0 \colon= \left\{ \ \frac 1 n \ \colon \ n \in \mathbb{N} \ \right\} \cup \{ 0 \}.$$
Finally let $$A \colon= \cup_{n=0}^\infty A_n.$$
Is this set $A$ good enough?
Your set $A$ is good enough. Your answer may not be -- you probably need to show that $A$ is closed and bounded and has a countable set of limit points.
The bounded part is easy, since every point in $A$ lies on the (closed) unit interval.
The closed part would be easy if you showed that the only limit points of $A$ are at zero and $\frac{1}{n}:n\in\Bbb{N}$. And that would also make the demonstration that the limit points are countable easy. So you have to show that these are the only limit points. A hint about how to do this: Can you prove that $\min |x - y| : x\in A_m, y \in A_n$ is some function of $m$ and $n$? What does this say about the possibility of a limit point of the union of the $A_i$ that is not a limit point of any particular $A_i$. Careful here, because zero is just such a limit point.
One final comment -- since any finite set is countable, an easier example would have been $\{ \frac{1}{n} : n\in\Bbb{N}$. However, Rudin's definition of countable is what we might call "countably infinite."