general topology - Prob. 13, Chap. 2 in Baby Rudin: Construct a compact set of real numbers whose limit points form a countable set

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?