Proofs with limit superior and limit inferior: $\liminf a_n \leq \limsup a_n$ I am stuck on proofs with subsequences. I do not really have a strategy or starting point with subsequences. 
NOTE: subsequential limits are limits of subsequences
Prove: $a_n$ is bounded $\implies \liminf a_n \leq \limsup a_n$ 
Proof:
Let $a_n$ be a bounded sequence. That is, $\forall_n(a_n \leq A)$.
If $a_n$ converges then $\liminf a_n = \lim a_n = \limsup a_n$ and we are done.
Otherwise $a_n$ has a set of subsequential limits we need to show $\liminf a_n \leq \limsup a_n$:
This is where I am stuck...

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
$$\{a_{n+1},a_{n+2},\cdots\}\subseteq \{a_n,a_{n+1},a_{n+2},\cdots\}$$
