I was hoping someone could look over my proof that every sequence has a monotonic subsequence. This is adapted out of Kenneth Ross's real analysis textbook. I am primarily considered about the logic I employ in constructing an inductive proof in order to construct an increasing subsequence, as I am a bit confused in what exactly the proposition I am proving at each step is, even though it does seem that the strategy of induction is used to construct such a sequence: for example, am I proving that I can create a subsequence each time, or am I showing that I can always find a successor term that is greater than or equal to the past term, and with the union of these resulting terms comprising the increasing subsequence? Most proofs I've seen just reference the principle of induction without writing the full proof out.
Consider the sequence $\left(s_n\right)$, and call some term, $s_m$, a dominant term if $s_m > s_n$ for all $n > m$. Consider two exhaustive cases.
Case 1: Suppose we have an infinite number of dominant terms. Call these terms $s_{n_1}$, $s_{n_2}$, $s_{n_3}, \ldots$. Then, $n_1 > n_2 > n_3 > \ldots > n_k$ implies that $s_{n_1} > s_{n_2} > s_{n_3} > \ldots > s_{n_k}$, and so we have a strictly decreasing, monotonic subsequence.
Case 2: Suppose we have only a finite number of dominant terms, meaning that these terms terminate. Call the final such dominant term $s_{n_0}$. Therefore, for all $N > n_0$, $\exists m > N$, $s_{m} \geq s_N$: put differently, for all terms beyond $s_{n_0}$, there exists some later term in the sequence that is at least as large as it.
We will proceed by induction to demonstrate how we can construct an increasing monotonic subsequence.
Base Case: Let $N = n_1$ and consider the term $s_{n_1}$. Since $n_1 > n_0$, $s_{n_1}$ is not a dominant term, and so there exists some $s_m$ such that $s_m \geq s_{n_1}$. Call this term $s_{n_1 + k_1}$ for some natural number $k_1$, so $s_{k_1} \geq s_{n_1}$.
Induction Hypothesis: Suppose that, with $N = n_k$, that we can construct some term beyond $s_{n_k}$ with the property that this term is at least as large as $s_{n_k}$. Call this term $s_{n_k + k_2}$, and so we have $s_{n_2 + k_2} \geq s_{n_k}$.
Induction Step: Let $N = n_{k+1}$. Since $n_{k + 1} > n_0$, there exists some element in the sequence at least as large as $s_{n_{k+1}}$. Call this term $s_{n_{k+1 + \alpha}}$, So, we have $s_{n_{k+1 + \alpha}} \geq s_{n_{k+1}}$.
So, for any arbitrary term, we can find some other term at least as large as it, and can therefore construct a monotonically increasing subsequence.
How does this look?