= Solution
{c}
Choose $M$ with $|b_n|\leq M$. Every <tail of a sequence> is nonempty and bounded, so its <supremum> $c_n$ exists and lies in $[-M,M]$. Removing one term cannot increase a <supremum>, hence $c_{n+1}\leq c_n$. The <bounded monotone sequence theorem>, applied to $-c_n$, proves
$$
c_n\longrightarrow L=\inf_n c_n=\limsup_{n\to\infty}b_n.
$$
Thus $L$ is the <limit superior> of $(b_n)$.
For $\varepsilon>0$, eventually $c_n<L+\varepsilon$, and then $b_n\leq c_n<L+\varepsilon$. Only finitely many indices can satisfy $b_n\geq L+\varepsilon$. If only finitely many satisfied $b_n>L-\varepsilon$, some entire <tail of a sequence> would have $b_n\leq L-\varepsilon$. Its <supremum> would also be at most $L-\varepsilon$, contradicting $c_n\geq L$.
Inductively choose $n_k>n_{k-1}$ so late that $c_{n_k}<L+1/k$ and $b_{n_k}>L-1/k$. There are infinitely many choices for the latter inequality, so both requirements can be met. Then $|b_{n_k}-L|<1/k$, proving $\boxed{b_{n_k}\to L}$. This constructs a <subsequence> attaining the <limit superior>.
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