Proposition 1.3.18 (Conway's A First Course in Analysis): If $\{a_n\}$ is a sequence in $\Bbb{R}$, $\alpha = \limsup a_n$, and $\beta = \liminf a_n$, then there is a subsequence of $\{a_n\}$ that converges to $\alpha$ and another that converges to $\beta$.
Here's the proof Conway furnishes for the reader:
I think the proof has a gap, since nothing Conway said guarantees that $\{n_k\}$ is a strictly increasing sequence in $\Bbb{N}$. Here's an attempt at fixing the gap:
Given $k=1$, since $\sup_{\ell \ge n} a_\ell$ is a sequence decreasing to $\alpha,$ there is some $N_1 \in \Bbb{N}$ such that $\sup_{\ell \ge N_1} - \alpha < 1$. Moreover, by the definition of supremum, there is some $n_1 \ge N_1$ such that $\sup_{\ell \ge N_1} a_\ell < a_{n_1} + 1$. Again, given $k=2$, there is some $N_2 \in \Bbb{N}$ such that $\sup_{\ell \ge n} a_\ell -\alpha < \frac{1}{2}$. If $N_2 \le n_1$, replace $N_2$ by $n_1 +1$; otherwise, just keep that index. By definition of the supremum, there is some $n_2 \ge N_2 > n_1$ such that $\sup_{\ell \ge N_2} a_\ell < a_{n_2} + \frac{1}{2}$. Continuing process (i.e., doing the induction which I'm too lazy to include), we obtain a strictly increasing sequence $\{n_k\} \subseteq \Bbb{N}$ such that...
Playing with the inequalities, we can show that $|\alpha - a_{n_k}| < \frac{1}{k}$, as Conway did. Does the above look right?
Conway's proof shows that there exists a sequence $\lbrace n_k \rbrace_{k \in \mathbb{N}^*}$ such that $a_{n_k} \rightarrow \alpha$. Moreover, by construction, you have $n_k \rightarrow +\infty$. So you can extract from this sequence to get a strictly increasing sequence $\lbrace n_{k_i} \rbrace$, and you have $a_{n_{k_i}} \rightarrow \alpha$.