I assume the elements of the sequence are real numbers. Note that for every positive integer $k$, if a segment of length $k+1$ contains $a_k$, then it contains $a_{k+1}$ as well. Hence $a_k\ne a_{k+1}$ and the sequence $b_1,b_2,\ldots\in\{-1,1\}$ defined by $b_n\doteqdot sign(a_{n+1}-a_n)$ is well-defined. It follows that for each positive integer $k$, the element $b_k$ is contained in a constant segment of length $k$. Moreover, it is equivalent to prove that the sequence $\{b_n\}$ is eventually constant. The main claim is as follows. Claim. Let $m$ be a positive integer such that $b_m\ne b_{m+1}$. Then $b_1=b_2=\ldots=b_m$. Proof. As proved above, $b_m$ is contained in a constant segment of length $m$. Since $b_m\ne b_{m+1}$, this segment cannot contain $b_{m+1}$. Hence this segment must be $b_1,\ldots,b_m$. $\square$ It follows that there is at most one positive integer $m$ such that $b_m\ne b_{m+1}$, so $\{b_n\}$ is eventually constant.