TITLE: Can a sequence have infinitely many limits among its subsequences?
QUESTION [41 upvotes]: Suppose we have an extended real (countably) infinite sequence $(x_n)$. Then consider all of its possible subsequences $(x_{n_k})$. We could then consider the set $$A = \{a\in \overline{\mathbb{R}}:x_{n_k}\rightarrow a \text{ for some subsequence}\}.$$
Must this set necessarily be finite?
Otherwise we have countably many numbers that the original sequence approaches arbitrarily closely countably many times in its tail. Is there some kind of argument that this would require an uncountable sequence? 
My thought is we could take the supposed countable set, and then choose some $\epsilon>0$ so that no two limits live in the same $\epsilon$-ball. Far enough down each subsequence, no element in the tail of one subsequence can also be in the tail of another. From here, is the argument like Cantor's diagonalization?
Limits: subseq
$a_1 : x_{n_1}, x_{n_2}, x_{n_3}, \dots$
$a_2 : x_{m_1}, x_{m_2}, x_{m_3}, \dots$
$a_3 : x_{o_1}, x_{o_2}, x_{o_3}, \dots$
We would need uncountably many limits for these subsequences to be unique, and this contradicts the original sequence being countable.

REPLY [4 votes]: Let $\{x\}$ denote the fractional part of a real number $x$, i.e.,
$$\{x\} = x - \lfloor x \rfloor.$$
Since $\pi$ is an irrational number, the sequence
$$\{a_n\} = \{\{n\pi\}: n = 1, 2, 3, \ldots\}$$
is dense in $[0, 1]$. (for the proof, see here.) Therefore, every point of $[0, 1]$ is a limit of some subsequence of $\{a_n\}$. 
This example provides an example that the set of limits of subsequences can be made uncountably infinite.