Let $\{a_n\}$ be a bounded sequence. Then we define the sequences $\{a_n^+\}$ and $\{a_n^-\}$ by
$$a_n^+=\sup\{a_n,a_{n+1}\dots\}$$
$$a_n^-=\inf\{a_n,a_{n+1}\dots\}$$
We (may) then define
$$\lim a_n^+=\limsup a_n$$
$$\lim a_n^-=\liminf a_n$$
Now, you need two things to work this out:
$(1)$ Let $A$ be any bounded nonempty subset of $\Bbb R$. Then
$$\inf A\leq \sup A$$
$(2)$ Let $\{\alpha_n\}$ be a sequence such that $a_n\geq0 $ for each $n\in \Bbb N$. Then $$\lim a_n\geq 0$$
With $(1)$ you should show $$a_n^-\leq a_n^+$$ for each $n\in \Bbb N$. Monotone convergence says both $\{a_n^+\}$ and $\{a_n^-\}$ converge, since they are bounded (above/below) and are monotone (increasing/decreasing)$^{(*)}$. But
$$a_n^+- a_n^-\geq 0$$
for each $n\in \Bbb N$, so use $(2)$ to show
$$\lim a_n^+-\lim a_n^-\geq 0$$
that is:
$$\liminf a_n\leq \limsup a_n$$
$(*)$ To prove this, you need to show that if $A\subseteq B$, then $$\sup A\leq \sup B$$ $$\inf A\geq \inf B$$ Then, observe that
$$\{a_{n+1},a_{n+2},\cdots\}\subseteq \{a_n,a_{n+1},a_{n+2},\cdots\}$$
HINT: Suppose that $X_n\le Y_n$ for each $n\in\Bbb N$, and suppose, to get a contradiction, that $\liminf_nY_n<\liminf_nX_n$. Then there is a subsequence $\langle Y_{n_k}:k\in\Bbb N\rangle$ that converges to some $y<\liminf_nX_n$. Show that either $\langle X_{n_k}:k\in\Bbb N\rangle\to-\infty$, or $\langle X_{n_k}:k\in\Bbb N\rangle$ has a subsequence that converges to some $x\le y$; both are impossible, since $y<\liminf_nX_n$.
The other inequality can be proved in a very similar way.
Best Answer
Here is how to prove $\liminf \{ x_n \} \leq \limsup \{ x_n \}$ for a general real sequence $\{ x_n \}$. Set $L = \liminf \{ x_n \}$ and $S = \limsup \{ x_n \}$, and suppose for the sake of contradiction that $L > S$. Say $L = S + h$ for some $h > 0$. Then there are infinitely many $n$ for which $x_n \in (L - h, L + h)$. Therefore, there are infinitely many $n$ for which $x_n > S$, contradicting the definition of $S$. Hence, $L \leq S$.