TITLE: When does convergence of subsequences implies convergence of sequence?
QUESTION [8 upvotes]: Let $X$ be a compact metric space and $(a_n)_{n \in \mathbb{N}}$ be a sequence in $X$ and $(a_{n_i})_{i \in \mathbb{N}}$ be a convergent subsequence (we know that such exists because $X$ is compact). Denote by $l$ its limit. My question is:

Suppose that, for every integer $k \geq 0$, the subsequence $(a_{n_i+k})_{i \in \mathbb{N}}$ of $(a_n)$ converges to $l$. Is it necessary that $(a_n)$ also converges to $l$?

I think that this is true, but I could not prove it playing around with the $\epsilon - \delta$ definition of limit. Any help?
Thank you in advance!

REPLY [3 votes]: Your conjecture is false.
Construct the original sequence in batches, where the $n$th batch consists of 
n zeroes followed by a single 1, that is, the sequence is 010010001... Suppose now your subsequence picks off the first zero in each batch. For each fixed k the translated subsequence will eventually be zero, but the sequence itself does not converge.