Let $(a_n)$ be a bounded sequence such that $\inf_{\ell} a_{\ell}<a_n< \sup_{\ell} a_{\ell}$ for each $n=1,2, \dots$
I want to show that there are subsequences $(a_{k_n})$ increasing and $(a_{m_n})$ decreasing such that $a_{k_n} \to \sup_{\ell} a_{\ell}$ and $a_{m_n} \to \inf_{\ell} a_{\ell}$.
Could you give me a hint how we could find the desired subsequences?
Hint
Suppose the infimum is $0$ and the supremum is $1$. Choose $a_{k_1}=a_1$ as the fist member of the decreasing sequence. Then since $a_1>0$ by hypothesis, there is an element $a_{k_2}$ with $k_2>k_1$ and $0<a_{k_2}<a_{k_1}/2$ and a $k_3>k_2$ and $0<a_{k_3}<a_{k_2}/2$ and so on. (Make sure you see why we can choose $k_{m+1}>k_m.$)
EDIT
In response to the OP's comment. First of all, since $0 = \inf{a_n}$ there are $a_k$ arbitrarily close to $0$ so we can certainly find $k_2$ such that $0<a_{k_2}<a_{k_1}/2$. The only question is whether we can find such a $k_2>k_1$. Suppose not. Then $K=\{k|a_k<a_{k_1}\}$ is finite, and then $\inf{a_k}=\min_{k\in K}a_k>0,$ contradiction.