A sequence converges if every subsequence has a sub-subsequence which converges to the same elt. Actually, I found the same questions in this website and saw the proof. 
What I'm looking for is somewhat intuitive understanding.
I know that, {$x_n$} converges to {$x$} if and only if every subsequence {$x_{n_k}$} converges to {$x$}.
So, if {$x_n$} converges to {$x$}, it looks trivial that every subsequence has a sub-subsequence that converges to {$x$}. Actually all the sub-subsequence converges to the same point, I think.
However, the opposite way is not easy for me. How can the existence of a convergent sub-subsequence in each subsequence guarantees the convergence of sequence?
For me, it looks like that in each subsequence, the existence of one sub-subsequence which converges guarantees the convergence of all sub-subsequence.
What am I missing? 
 A: Suppose your sequence does not converge to $x$. Therefore, there exists a neighbourhood of $x$ which, no matter how big $N$, there is always some $x_n$ for $n>N$ such that $x_n$ is not on that neighbourhood. This way, you can construct a subsequence which never enters that neighbourhood. But this subsequence will not be able to have any subsequence converging to $x$, by its construction.
A: There is really no need for a proof of any type if you know that any convergent sequence shares its limit with any subsequence. Let $A$ be a convergent sequence and $B$ be any subsequence of $A$. Let $C$ be any subsequence of $B$. Then $B$ and $C$ share the same limit, therefore so does $C$ and $A$ and you are done. The main thing is that a subsequence of a subsequence of a sequence is a subsequence of the original sequence. That is a bit of a mouthful, but try to understand that statement. 
A: It is because the sequence itself is an actual subsequence of itself. 
So for all the subsequences to converge, the actual sequence has to converge to.
(and similarly for a sub-subsequence). 
