Sufficient condition for convergence of a real sequence Let $(x_n)$ be a sequence of real numbers.
Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the original sequence $(x_n)$ itself converges to $x$ .            
Thanks for any help.
 A: Suppose $x_n$ does not converge to $x$, but every subsequence of $x_n$ has a sub-subsequence which converge to $x$.
Since $x_n$ does not converge to $x$ we must be able to find a subsequence such that every term is more than $\epsilon$ away from $x$ for some $\epsilon>0$, but clearly this does not have a sub-subsequence which converges to $x$, by definition.
A: Let every subsequence of $ x_n$   has a convergent subsequence to $ x$  and suppose by way of contradiction that $x_n$ does not converges to $x$  . Then there exists $ ε>0$  such that for every $ n_0$ , $ |x-x_n|\geq ε$  for some  $n\geq n_0$. Thus $ |x-x_n|\geq ε$ for an infinite number of $n$ . This implies that there exists a subsequence $ y_n$  of $ x_n $ , such that for each $n$, $ |x-y_n|\geq ε $ . However the latter contradicts the fact that $ y_n$  has a subsequence that converges to $ x$  . 
A: First notice that your condition implies that your sequence is bounded.
Indeed, if $(x_n)$ is unbounded, we can find a subsequence $(x_{n_k})$ such that $|x_{n_k}|\ge k$. This subsequence does not have a convergent subsequence.

So we know that $(x_n)$ is bounded and it is not convergent. 
This means that $$M=\limsup x_n > \liminf x_n =m.$$
(Both $M$ and $m$ are real numbers, since $(x_n)$ is bounded.)
We know (from the properties of limit superior and limit inferior) that there is a subsequence $(x_{n_k})$ which converges to $M$ and there is a subsequence $x_{n_l}$ which converges to $m$. (And every subsequence of any of these two subsequences has, of course, the same limit $M$ resp. $m$.)
We have found two subsequences with different limits, which contradicts your assumptions about the sequence $(x_n)$.
