= Solution
A real <sequence> has limit $a\in\mathbb R$ if, for every $\varepsilon>0$, there is $N$ such that $|a_n-a|<\varepsilon$ for all $n\ge N$. It tends to $+\infty$ if, for every real $M$, eventually $a_n>M$; it tends to $-\infty$ if eventually $a_n<M$ for every real $M$.
The <Bolzano-Weierstrass theorem> states that every <bounded sequence> of real numbers has a real convergent <subsequence>. This proves the claim for bounded sequences. If the sequence is unbounded above, every tail remains unbounded above because a finite prefix is bounded. Choose indices $n_j$ increasing with $a_{n_j}>j$. This subsequence tends to $+\infty$. If it is unbounded below, choose $a_{n_j}<-j$ to obtain a subsequence tending to $-\infty$. This proves the <extended real subsequence theorem>.
For the final example, a <bounded sequence with vanishing increments need not converge>. Take
$$
\boxed{a_n=\sin\sqrt n.}
$$
The <mean value theorem> gives
$$
|a_{n+1}-a_n|\le\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n}\longrightarrow0.
$$
Nevertheless, along $n_j=\lfloor(\pi/2+2\pi j)^2\rfloor$, the square-root rounding error tends to zero, so $a_{n_j}\to1$. Along $m_j=\lfloor(3\pi/2+2\pi j)^2\rfloor$, it tends to $-1$. These two <subsequences> have different limits, proving nonconvergence.
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