While typically, the Axiom of Choice and its peripheral arguments are not emphasized in one's first exposure to Real Analysis, I am trying to be as rigorous as possible in my learning as an axiomatic set theory enthusiast. As per the question, is the Axiom of Countable Choice sufficient in this proof? My main concern is that I specified $n_{k + 1} > n_k$ in the construction, but I think each $E_m$ being infinite should guarantee that each term exists. If the ADC is not needed here, in what cases is it necessary?
Lemma 1: A monotonically increasing (decreasing) unbounded sequence tends to $\infty$ ($-\infty$).
Proof. Let $(s_n)$ be such a sequence. Fix $M > 0$. Then there existsan $N \in \mathbb{Z^+}$ such that $s_N > M$, as otherwise $(s_n)$ would bebounded above by $s_N$ and below by $s_1$, hence bounded. It is clearthat the $N$-tail of $(s_n)$ is contained in $(M, \infty]$, asrequired.
Theorem. If a sequence $(s_n)$ in $\mathbb{R}$ is not bounded above (below), then it has a subsequence $(s_{n_k})$ such that $s_{n_k} \to \infty$ ($-\infty$).
Let $E_m = \{s_n : s_n > m\}$ for every $m \in \mathbb{Z^+}$. Each $E_m$ is infinite, as otherwise there exists an $M \in \mathbb{Z^+}$ such that $\max(E_m)$ is an upper bound of $(s_n)$. The axiom of countable choice then specifies a subsequence $(s_{n_k})$ such that $s_{n_k} \in E_k$ and $n_{k + 1} > n_k$. Hence, $(s_{n_k})$ is monotonically increasing and unbounded above, and the theorem follows from the lemma.