Cluster points and the sequence 1,1,2,1,2,3,1,2,3,4,1,... I am working on a problem in analysis.  We are given a sequence $x_n$ of real numbers.  Then a definition: A point $c \in \mathbb{R}\cup{\{\infty, -\infty}\}$ is a cluster point of $x_n$ if there is a convergent subsequence of $x_n$ with limit $c$.  Then, we let $C$ bet the set of all cluster points.  The problem is to prove that the set $C$ is closed.  In working on this problem, I tried to determine the cardinality of $C$.  And, in doing this I came up with the sequence $x_n = 1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,...$ as an example of a sequence with countable number of cluster points.  I believe that for this sequence $C = \mathbb{N}$.
Now consider an enumeration of $\mathbb{Q}$, $q_1, q_2, q_3,...$  And define the sequence $y_n = q_1, q_1, q_2, q_1, q_2, q_3, q_1, q_2, q_3, q_4,...$  It seems to me that the set of cluster points of this sequence is $\mathbb{Q}$.
But now, we could take a sequence from the cluster points of $y_n$ (which is $\mathbb{Q}$) that converges to a member of $\mathbb{Q}^c$ (like $\sqrt{2}$), which would contradict the assumption that the set of cluster points is closed.  (Since we have a theorem: a subset of a metric space is closed iff every convergent sequence in the set converges to an element of that set).
So, in trying to prove $C$ is closed, I have come up with what I believe is a counterexample.  So my question is: where is the error?

For the record, I would like to suggest a topological answer to prove $C$ is closed. Note that $\mathbb{N}$ is closed hence a limit point $x$, of a sequence $f:\mathbb{N}\rightarrow\mathbb{N}$ is in $\mathbb{N}$. The set of cluster points $C$ is equal to the limit points of $f$ restricted to all infinite subsets of $\mathbb{N}$, hence $C\subset\mathbb{N}$ and any subset of $\mathbb{N}$ is closed in $\mathbb{R}$.
Note the same proof does not hold for $f:\mathbb{N}\rightarrow \mathbb{Q}$, since $\mathbb{Q}$ is not closed.
Note the same proof does not hold for $f:\mathbb{N}\rightarrow \mathbb{R}$, since $\mathbb{R}$ does not have a discrete topology.
