Q: Let $(x_n)$ be a real sequence satisfying that every subsequence of $(x_n)$ does not converge in $\Bbb{R}$. Prove that $\vert{x_n}\vert$ $\implies$ $\infty$ as $n$ $\implies$ $\infty$.
My lecturer left this as an exercise that he didn't give the answer to, he set up the proof as such.
Proof: By contradiction, if $\vert{x_n}\vert$ does not $\implies$ $\infty$ (negation of $\vert{x_n}\vert$ $\implies$ $\infty$). Then $\exists$$M$ $>$ $0$, such that $\forall$ $N$ $\in$ $\Bbb{N}$, such that $\vert{x_{n^{'}}}\vert$ $\leq$ $M$
His words are "From this, there is some more work by induction, say; this will imply that there is a subsequence $(x_{n_{k}})$ using this method we can deduce a subsequence such that $(x_{n_{k}})$ $\leq$ $M$
He then continues, as $(x_{n_{k}})$ is a bounded sequence. By Bolzano-Weierstrass Theorem, there's a convergent subsequence of $(x_{n_{k}})$, call it $(x_{n_{k_{j}}})$. This is a contradiction as $(x_{n_{k_{j}}})$ is a subsequence of the original sequence $(x_n)$ which converges in $\Bbb{R}$
I just want help with the induction proof for the implication that the subsequence $x_{n_{k}}$ is bounded from the middle. How would I do this? How would this be done? I'm not sure.
