Claim
Set of Cluster(Accumulation) points of a sequence $x_k$ is closed.
Proof
A set is closed if and only if its accumulation(limit) point in that set.
So we can change the claim into "set of cluster points of a sequence $x_k$ holds the limit points (of the set of cluster points of a sequence)".
On
Denote the set of cluster points of a set $E$ by $E'$.
If $\{x_n\}''$ is empty, then $\{x_n\}'$ is closed.
Suppose $x \in \{x_n\}''$. Let $U$ be an open set containing $x$.
By definition, $U$ contains infinitely many points of $\{x_n\}'$. Let $y$ be one such point.
Then $U$ is an open set containing $y \in \{x_n\}'$ so again by definition $U$ contains infinitely many points of $\{x_n\}$.
To recap, any open set containing $x$ contains infinitely many points of $\{x_n\}$ so that $x \in \{x_n\}'$.
Let $S$ denote the set of elements that are not clusterpoints of sequence $(x_n)_n$ and let $s\in S$.
By definition an open set $U_s$ exists such that $\{x_n\in U_s\mid n\in\mathbb N\}$ is finite.
But actually this shows that every element of $U_s$ is not a clusterpoint of the sequence.
So we have $U_s\subseteq S$.
Now observe that $S=\bigcup_{s\in S}U_s$ which - as union of open sets - is open itself.
Then its complement, i.e. the set of clusterpoints of sequence $(x_n)_n$ is closed.