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.
Best Answer
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 $x$?