For any closed subset of $\mathbb{R}$ there is a sequence in $\mathbb{R}$ whose sequential limits is equal to the that subset Question: Let $A$ be a closed subset in $\mathbb{R}$. Prove that there exists a sequence $x_n$ in $\mathbb{R}$ whose set of subsequential limits is exactly equal to $A$. 

My approach:
I think this sort of makes sense me if we our space was $\mathbb{C}$, however, I'm not even sure if this statement is true or not. I was thinking if it's possible to prove it by starting to say that the set of all subsequential limits of a sequence in a metric space is closed. However, I'm not sure if for any closed subset in $\mathbb{R}$, we can construct such sequence.
I'm going to assume that you know basic things about compactness. And give you two steps to begin with.


If $A$ is a closed and bounded set, then it is compact. Prove, even in general, that a compact metric space is always the closure of a countable set (or in other words, a compact metric space is always countable). You can do that by finding a particular sequence of open covers, refine each to a finite subcover and use that to define the countable set.
If $A$ is closed and unbounded, then it can be written as the countable union of closed and bounded sets. Namely, if $A$ is closed, then it is the union of countably many compact sets. What do you know about the countable unions of countable sets? Now apply the previous case.


Or, prove directly,


If $A\subseteq\Bbb R$ is covered by open sets, then there is a countable subcover. Use similar tricks to the previous two steps and find a countable dense subset. (This property of $\Bbb R$ is called hereditarily Lindelöf.)
Hint: $A$ contains a countable dense subset.