It is usually stated that any sequence of real numbers $\big( a_n\big)_{n=1}^{\infty}$ has either an increasing or a decreasing subsequence. If $(a_n)$ is not bounded, wlog say it has no upper bound, then it contains an increasing subsequence that converges to $+\infty$. So, suppose $(a_n)$ is bounded. Then it has a limit point, say $a$. It means, any neighborhood $(a-\varepsilon,a+\varepsilon), \varepsilon>0$ of $a$ contains infinitely many terms of $(a_n)$. Hence, at least one of the following two options holds. 1) For each $\varepsilon>0$ there are infinitely many terms of the sequence $(a_n)$ contained in $[a,a+\varepsilon)$. 2) For each $\varepsilon>0$ there are infinitely many terms of the sequence $(a_n)$ contained in $(a-\varepsilon, a]$. In the first case we have a decreasing subsequence of $(a_n)$ that converges to $a$, and if the second one holds, there is an increasing subsequence of $(a_n)$ that converges to $a$.