Repeating the question from the title:
Question. Let $(a_i)_{i=1}^{\infty}$ be a sequence. How to show that $\liminf_{n \rightarrow \infty} ≤ \limsup_{n \rightarrow \infty}$?
So far I've only managed to show that the inequality holds in one specific case, that is, the case where $\limsup_{n \rightarrow \infty}$, $\liminf_{n \rightarrow \infty}$ are the elements of sequences $(a_{i}^{+})_{i=1}^{\infty}$ and $(a_{i}^{-})_{i=1}^{\infty}$, respectively (in this case the proposition holds, since for arbitrary $N,M ≥ 1$, we have $\sup(a_i)_{i=N}^{\infty} ≥ \inf(a_i)_{i=M}^{\infty}$, even if $N ≠ M$).
I've tried to generalize by using contradiction, but it led me nowhere. More concretely, if for the sake of contradiction we assume that $\liminf_{n \rightarrow \infty} > \limsup_{n \rightarrow \infty}$, then it must be the case that $\liminf_{n \rightarrow \infty} > a_n$ for all $n≥N$ and some $N$ (this is the result from the previous exercise in the textbook). This doesn't lead to any contradiction though, since there are sequences where limit inferior is larger than all elements of the sequence past a specific point (e.g., consider the sequence $0.9,0.99,0.999,\cdots$; limit inferior in this case is $1$, which is larger than any element of the sequence)