A question on a proof that every sequence has a monotone subsequence

If you have a statement "for all $x$, $P(x)$ is true", then the negation of this statement is "there exists an $x$ such that $P(x)$ is not true". (the negation of $\forall x: P(x)$ is $\exists x: \neg P(x)$).

$n_1$ is not a peak. The definition of $n_1$ being a peak is:

$$\forall \,n> n_1: x_{n_1}>x_n$$ meaning that by definition, $n_1$ is not a peak iff $$\exists \, n > n_1: x_n\leq x_{n_1}$$ which is exactly what you have.

Here is a tedious proof that avoids dealing with peaks:

Let $\bar{x} = \limsup_n x_n$.

If $\bar{x} = \infty$, then let $n_1 = 1$, and $n_{k+1} = \min \{n | n \ge n_k+1,\ x_n \ge x_{n_k} +1 \} $, then $x_{n_k}$ is a monotone sequence.

Otherwise, suppose $\bar{x} < \infty$.

Suppose $\{x_n\} \cap (\bar{x},\infty)$ contains an infinite subset. Let $x_{n_k}$ be the subsequence of $x_n$ that lies in $(\bar{x},\infty)$. Then $x_{n_k} > \bar{x}$, and $\lim_k x_{n_k} = \bar{x}$. Let $m_1 = n_{1}$, and $m_{k+1} = \min \{n_j | n_j \ge m_k+1,\ x_{n_j} < x_{m_k} \} $, then $x_{m_k}$ is a monotone sequence.

Now suppose $\{x_n\} \cap (\bar{x},\infty)$ is finite.

Suppose $\{x_n\} \cap \{\bar{x}\}$ contains an infinite subset, then clearly this subsequence is a monotone (in fact, constant) subsequence.

Now suppose $\{x_n\} \cap [\bar{x},\infty)$ is finite. Then $x_n < \bar{x}$ (after a finite number of terms), and there exists a subsequence $x_{n_k}$ such that $\lim_k x_{n_k} = \bar{x}$. Similar to above, let $m_1 = n_1$, and $m_{k+1} = \min \{n_j | n_j \ge m_k+1,\ x_{n_j} > x_{m_k} \} $, then $x_{m_k}$ is a monotone sequence.