I know the issue has already been addressed in other posts, but I want to ask if the proof I constructed is correct - I'm not confident with proofs.
If the sequence (xn)n is unbounded, then by definition
∀M>0, ∃n∈$\mathbb{N}$ s.t.|xn| $\geq$ M.
To construct a subsequence (xkn)kn that tends to plus infinity select the stricly increasing sequence of natural numbers (kn)n such that (xkn)kn is increasing, thus ∀M>0, ∃n∈$\mathbb{N}$ s.t. xkn $\geq$ M.
To construct a subsequence (xkn)kn that tends to minus infinity select the stricly increasing sequence of natural numbers (kn)n such that (xkn)kn is decreasing, thus ∀M>0, ∃n∈$\mathbb{N}$ s.t. xkn $\leq$ M.
Is it strong enough? Do I miss something?
Your construction is not complete. It is merely restating what you have to prove. In fact if a sequence is unbounded there is no guarantee that there exists two subsequences, one tending to $\infty$ and the other to $-\infty$. Is $x_n$ is not bounded above choose $n_1$ such that $x_{n_1} >1$. Then choose $n_2$ such that $x_{n_2} >2$ and so on . By induction you get $\{x_{n_k}\}$ such that $x_{n_k}>k$ for all $k$ which implies $x_{n_k} \to \infty$. Use a similar argument if the given sequence is unbounded below.