= Solution
{c}
If $a_n\to L$, every <subsequence> also has <limit of a sequence> $L$: its increasing indices eventually exceed any prescribed threshold. This applies in particular to every <monotone subsequence>.
Conversely, suppose $(a_n)$ is a <bounded sequence> and all its <monotone subsequences> converge to the same $L$. At least one exists by the <monotone subsequence theorem>. If $a_n$ did not tend to $L$, there would be $\varepsilon_0>0$ and a <subsequence> $(a_{n_j})$ such that
$$
|a_{n_j}-L|\geq\varepsilon_0\qquad(j\geq1).
$$
Apply the <monotone subsequence theorem> to $(a_{n_j})$. The resulting further <subsequence> is also a <monotone subsequence> of $(a_n)$, so it tends to $L$, contradicting the displayed bound. Therefore
$$
\boxed{a_n\to L\ \Longleftrightarrow\ \text{every monotone subsequence tends to }L.}
$$
The <bounded monotone sequence theorem> ensures that these <subsequences> have finite <limits of sequences>; the hypothesis requires those values to agree.
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