Construct a compact set of real numbers whose limit points form a countable set.

What about $$A=\left\{\frac1n+\frac1 m:m,n\in\Bbb N\right\}\cup\{0\}\text{ ? }$$

One can see the $A'$ is $$\left\{\frac 1 n:n\in\Bbb N\right\}\cup \{0\}$$ Thus, let $E=A\cup A'=\bar A$ which is closed, and bounded.

The idea you used is good. I will do the same thing more slowly, building the set step by step in order to retain control over the geometry.

The backbone of the set is, like yours, the sequence $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$. This sequence has limit $0$, so we throw in the number $0$.

Now near $\frac{1}{n}$ for every $n$, we produce a sequence that has limit $\frac{1}{n}$, and that does not mix in unpleasant ways with other sequences. Note that the distance from $\frac{1}{n}$ to $\frac{1}{n+1}$ is $\ge \frac{1}{2n^2}$.

So for every $n$, throw in the numbers $\frac{1}{n}+\frac{1}{(2n^2)(2^m)}$, where $m\ge 1$.

Another way: Start with the set $A$ of numbers $n+\frac{1}{2^m}$, where $n$ and $m$ range over the positive integers. This set does the job beautifully. Only one minor flaw: the set is not compact. Let $B$ be the set of all reciprocals of numbers in $A$, together with $0$. This does the job.

The limit points are $\{\frac{1}{2^m}\mid m\in \mathbb{N}\}$. These are contained in the set (to get $\frac{1}{2^k}$ (for $k>1$), take $m=k-1$, $n=2$).

We can tell there are no other limit points, since the closest points to $\frac{1}{2^k}(1-\frac{1}{l})$ (for $l>2$) are $\frac{1}{2^k}(1-\frac{1}{l+1})$ and $\frac{1}{2^k}(1-\frac{1}{l-1})$, so we can isolate them in a neighborhood of radius $\frac{1}{2^k}(1 - \frac{1}{2(l+1)})$.

Edit: As Andre has pointed out, $1/2$ is not in the set, so the problem does not work as stated.

What about the trivial unitary set as in $\{ 0 \}$?