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It is useful to prove the contrapositive. Suppose $k$ is a peak of a sequence $(a_n)$ and $m$ is a peak of a sequence $(-a_n)$. The second condition says $a_j\ge a_m$ for every $j>m$. HenceOnly finitely many earlier terms remain, and their absolute values have a finite maximum. Together these give a bound for every term, so $(a_n)$ is a bounded sequence. Consequently,

Call $k$ a peak of a sequence when $a_k\ge a_m$ for every $m>k$. If there are infinitely many peaks of a sequence, list their indices in increasing order, $k_1<k_2<\cdots$. Since $k_{r+1}>k_r$,This is a decreasing subsequence; throughout this argument, a decreasing monotone sequence may have equal successive terms.

If there are only finitely many peaks of a sequence, choose $N$ beyond all their indices; take $N=1$ if there are none. Every $n\ge N$ fails to be a peak of a sequence, so some $m>n$ satisfies $a_m>a_n$. Starting with $n_1=N$, choose such an $n_{r+1}>n_r$ recursively. This givesThe two cases prove the monotone subsequence theorem without any boundedness hypothesis.

Choose the sequenceAfter every index there are positive terms arbitrarily large and negative terms arbitrarily small. Thus every term has both a larger later term and a smaller later term. The former excludes a peak of a sequence $(b_n)$, and the latter excludes a peak of a sequence $(-b_n)$.

Choose the sequenceEvery even index is a peak of a sequence $(a_n)$, since its value is $1$ and all later values are at most $1$. Every odd index is a peak of a sequence $(-a_n)$ for the same reason. Both sets of indices are infinite.

Enumerate both sets of peaks of a sequence in increasing order of index. If $a_n\to L$, the two selected subsequences both tend to $L$.

For the converse, suppose $a_{k_r}\to L$ and $a_{m_r}\to L$. Given $\epsilon >0$, choose $r$ large enough that both selected terms differ from $L$ by less than $\epsilon$. For every $n>\max \{k_r,m_r\}$, the two peak of a sequence conditions implyThe lower bound comes from the peak of a sequence of $(-a_n)$ at $m_r$. This is precisely the definition of $a_n\to L$. Thus convergence is equivalent to a common finite limit of a sequence for the upper and lower peak subsequences.

Choose $N$ beyond every peak of a sequence of both $(a_n)$ and $(-a_n)$. Every term with index at least $N$ has a strictly larger later term and a strictly smaller later term.

Suppose for a contradiction that $a_n\to L$. Starting at $a_N$, recursively select a strictly increasing subsequence and a strictly decreasing subsequence. Both must have limit of a sequence $L$. If their second terms are $u>a_N$ and $v<a_N$, respectively, passage to their limits of sequences givesa contradiction. Hence $(a_n)$ does not converge. This argument includes the possibility that either set of peaks of a sequence is empty.

A peak of a sequence of real numbers is an index $k$ whose value is at least every later value. This is a condition on the entire tail, stronger than comparing only adjacent terms. Equality is allowed: every index of a constant sequence is a peak of a sequence.

Infinitely many peaks of a sequence, listed in increasing order of index, give a nonincreasing monotone subsequence. If only finitely many exist, every index beyond the last has a larger later term; repeatedly choosing one gives a strictly increasing monotone subsequence. This proves the monotone subsequence theorem.

A peak of a sequence of $(-a_n)$ supplies a lower bound for all subsequent terms of $(a_n)$. If both $(a_n)$ and $(-a_n)$ have a peak of a sequence, the two selected values bound a common tail. Adding bounds for the finite initial segment proves that $(a_n)$ is a bounded sequence.