Construct a set of real numbers whose limit points from a countable set. Is the set you constructed closed? Is it compact?
My example is $$G=\{1/n+1/m: n, m \in \mathbb N\}\cup \{0\}$$ and as $m$ goes to infinity, $G'=\{1/n, n\in\mathbb N\}\cup \{0\}$ which is an infinite subset of countable set.
You're almost there; your example certainly works, and you correctly computed $G$ and $G'$. Moreover, you have that $G$ is closed and compact - however your proof that it is closed doesn't look quite right. It shouldn't be talking about limit points or anything like that. The statement you want to prove to show that $G$ is closed is: $$\{1/n:n\in\mathbb N\}\cup\{0\}\subseteq \{1/n+1/m:n,m\in\mathbb N\}\cup\{0\}$$ and it doesn't matter if something on the left side is a limit point of something one the right (in fact, that's trivial, because that's you defined $G'$). So a correct proof that $G$ is closed might proceed to note that $0$ is in $G$ and, for any element $\frac{1}n$, which is in $G'$, it can be written as $\frac{1}{2n}+\frac{1}{2n}$, which is the sum of the reciprocal of two integers and is hence in $G$. This establishes that $G$ is closed, which, as you proved yourself, shows $G$ is compact too.