Is my argument correct?
Proposition. There is no divergent monotone sequence with a Cauchy subsequence.
Proof. Assume that we have a divergent monotone sequence $(a_n)$ with a Cauchy sequence $(a_{n_k})$, from theorem $\textbf{2.6.3}$ we know that there exists an $M>0$ such that $|a_{n_k}|<M,\forall k\in\mathbf{N}$.
We now examine the case where $(a_n)$ is increasing, the case where $(a_n)$ is decreasing will be handled similarly. Let $r\in\mathbf{N}$, evidently $r\leq n_r$ but then $a_r\leq a_{n_r}$ and since $a_{n_r}\leq |a_{n_r}|\leq M$ consequently $a_r\leq M$, implying that $(a_n)$ is bounded, but $(a_n)$ cannot be bounded since this together with $(a_n)$ being increasing would imply $(a_n)$ is convergent by theorem $\textbf{2.4.1}$.
$\blacksquare$
Note:
$\textbf{2.6.3}$ All Cauchy sequences are bounded.
$\textbf{2.4.1}$ All bounded monotone sequeces are convergent.