Convergence of sequence with subsequences Let ${a_n}$  be a bounded sequence such that any of its convergent proper subsequences converge to the same limit $l$. Prove that ${a_n}$ converges to $l$.
Any hints please? 
 A: Hint: Suppose $a_n$ does not converge to $l$. Then there is $\varepsilon \gt 0$ and an increasing sequence $n_k \in \mathbb{N}$ such that $|a_{n_k} - l| \gt \varepsilon$ for any $k \in \mathbb{N}$. By Bolzano-Weierstrass $a_{n_k}$ has a convergent subsequence. Show that the limit of this subsequence cannot be $l$ and use this to conclude.
A: The statement $L=\lim_{n\to \infty}a_n$ means $$\forall r>0\;\exists n\;\forall m>n\;(|a_m- L|<r).$$ 
So $\neg (L=\lim_{n\to \infty}a_n)$ means $$\exists r>0 \forall n\; \exists m,m'>n\;(|a_m-L|\geq r).$$ What does that $mean$? It means that there is an $r>0$  for which we can find 
$m_1>1$ with $|L-a_{m_1}|>r,$ and
$m_2>1+m_1$ with $|L-a_{m_2}|>r$ and 
$m_3>1+m_2$ with .....(etc).....  
The set $S=\{m_2,,m_3,m_4,...\}$   excludes  $1+m_j$ for every $j\in \Bbb N.$  Consider the sequence $(a_k)_{k\in S}\;$.
A: Since $a_{n} $ is bounded the numbers $\liminf a_{n}, \limsup a_{n} $ exist (finitely). And there are subsequences of $a_{n} $ which converge to these numbers. But then every convergent subsequence of $a_{n} $ converges to the same limit $l$ it follows that these two numbers are same and equal to $l$ and hence the sequence $a_{n} $ converges to $l$. 
A: If $a_n \not \to l$ then there's an $\epsilon > 0$ so that for each $k$ we may find an $n_k \ge k$ such that $|a_{n_k} - l| \ge \epsilon$.  Clearly, we may also take $n_1 < n_2 < \dots$.  So $(a_n)$ has one nonconvergent subsequence, which is the contrapositive of your statement
A: It would be a good start assuming that the sequence does not converge to $l$.
Then try to find a proper subsequence that does not converge to $l$.
