Consider the following proof that every sequence $\{x_n\}_n$ in $\mathbb R$ has a monotone subsequence.
Proof: Let us call a positive integer $n$ a peak of the sequence if $m > n \implies x_n > x_m$ i.e., if $x_n$ is greater than every subsequent term in the sequence.
Suppose first that the sequence has infinitely many peaks, $n_1 < n_2 < n_3 < … < n_j < …$. Then the subsequence $\{x_{n_j}\}_j$ corresponding to these peaks is monotonically decreasing, and we are done.
So suppose now that there are only finitely many peaks, let $N$ be the last peak and set $n_1 = N + 1$.
Then $n_1$ is not a peak, since $n_1 > N$, which implies the existence of an $n_2 > n_1$ with $x_{n_2} \geq x_{n_1}.$ Again, as $n_2 > N$ it is not a peak, hence there is $n_3 > n_2$ with $x_{n_3} \geq x_{n_2}.$ Repeating this process leads to an infinite non-decreasing subsequence $x_{n_1} \leq x_{n_2} \leq x_{n_3} \leq \ldots$ as desired.
In the second case, if $n_1$ is not a peak, how does this imply the existence of an $n_2 > n_1$ with $x_{n_2} \geq x_{n_1}$? Can't it be possible that there does not exist any $n_2 > n_1$ such that $x_{n_2} \geq x_{n_1}$?
A possible example can be the sequence $\{\sin n\}_n$, taking $N$ to be the closest integer to $\pi/2$