I'm aware that other posts have discussed the same issue. I would like feedback on my specific proof. Thank you :)
Consider any sequence $(x_n).$
A "peak" element is an element such that all elements $x_i$ that succeed it in the sequence are no greater than it. In other words, an element $x_p$ is a "peak" element if for all $i > p_1,$ it is the case that $x_{p_1} \geq x_i.$
In the sequence $(x_n),$ either there are finitely many peaks, or there are infinitely many.
If there are finitely many, then there is a peak element $x_p$ that appears last. Then, for all $i \in \mathbb{N}$ such that $i > p,$ there exists some $M \in \mathbb{N}$ such that $M > i$ and $x_M > x_i.$ In such a case, there clearly exists a monotone increasing subsequence $(x_{n_i}).$
If there are infinitely many peaks, then there clearly exists a monotone decreasing subsequence,
$$x_{p_1}, x_{p_2}, x_{p_3}, \ldots$$
where $x_{p_1} \geq x_{p_2} \geq x_{p_3} \geq \cdots$ are all peak elements.
Hence, any sequence has some subsequence that is monotone, whether increasing or decreasing.
Consider any sequence $(x_n).$
A "peak" element is an element such that all elements $x_i$ that succeed it in the sequence are no greater than it. In other words, an element $x_p$ is a "peak" element if for all $i > p,$ it is the case that $x_{p} \geq x_i.$
In the sequence $(x_n),$ either there are finitely many peaks, or there are infinitely many.
If there are finitely many, then there is a peak element $x_p$ that appears last. Then, for all $i \in \mathbb{N}$ such that $i > p,$ there exists some $M \in \mathbb{N}$ such that $M > i$ and $x_M > x_i.$ In such a case, there clearly exists a monotone increasing subsequence $(x_{n_i}).$
If there are infinitely many peaks, then there clearly exists a monotone decreasing subsequence,
$$x_{p_1}, x_{p_2}, x_{p_3}, \ldots$$
where $x_{p_1} \geq x_{p_2} \geq x_{p_3} \geq \cdots$ are all peak elements.
Hence, any sequence has some subsequence that is monotone, whether increasing or decreasing.