If every subsequence has a further subsequence that converges to $x$, then the sequence converges to $x$ 
If every subsequence has a further subsequence that converges to $x$, then the sequence converges to $x$

In the proof, we assume the contrary, then there is some $\epsilon$ such that $|x_{n} -x|>  \epsilon$, $\forall n>N.$ 
Then we can find some subsequence which also doesn't converge to $x$. 
But then can't we still find a further subsequence which may converge to $x$? There certainly are further subsequence that don't converge, but it doesn't mean that all further subsequences don't converge? 
Please clarify the last part. 
 A: My experience was that this statement is easiest to prove by contraposition. In this case you want to prove that if $x_n$ does not converge to $x$ then there exists a subsequence $x_{n_k}$ such that all further subsequences $x_{n_{k_\ell}}$ do not converge to $x$. 
If we write the definition of $x_n \not \to x$, we have
$$(\exists \varepsilon > 0)(\forall N \in \mathbb{N})(\exists n \geq N) \, |x_n - x| > \varepsilon.$$
This definition allows us to extract a subsequence which is in fact bounded away from $x$: we fix such a $\varepsilon$, then we apply the universal for each $N \in \mathbb{N}$, which must necessarily give us countably many $n$ with the desired property. Why does this subsequence not have a further subsequence which converges to $x$?
A: Al right I think I see whats your problem. When you asume $x_n \not\rightarrow x$, what you are saying is:
There exist some $\varepsilon > 0$ such that for all $n \in \mathbb{N}$ (you fix an $n$, show an $n_k$, fix another $n$ show another $n_k$, etc) we have some $n_k >n$ such that $|x_{n_k} - x| \geq \varepsilon$. In simpler terms what you get is a subsequence such that $|x_{n_k} -x| \geq \varepsilon$ for all $n_k$ and this subsequence can't converge by definition. 
Remembering the definition of convergence: $x_n \rightarrow x$ means that given an arbitrary $\varepsilon' > 0$, there exists some $N$ such that for all $n \geq N$, $|x_n -x |< \varepsilon'$. 
Now if you take $\varepsilon' < \varepsilon$ you would see that $x_{n_k} \not\rightarrow x$ because every $x_{n_k}$ is at distance $\varepsilon$ from $x$ wich is greater than $\varepsilon'$
A: For a sequence $x_n$ define $\alpha = \limsup x_n$ and $ \beta = \liminf x_n $, which always exist, and for simplicity assume that they are finite. Consider the subsequence that converges to $\alpha$, denote it by $x_{n_k}$. Then this subsequence has a further subsequence that converges to $x$ by definition. But since the limits are unique we must have $x=\alpha$. Similarly, $\beta = x$. Thus
$$\liminf x_n = \limsup x_n = x $$
And so the sequence converges to $x$.
