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...

To make @Pedro's answer more complete, here's how things work in the unbounded case. Now we're given any sequence $\{a_n\}$, and would like to show $\liminf a_n \leq \limsup a_n$. I use the more general definition: $\liminf a_n := \sup_N a_N^- = \sup_N \inf_{n\geq N} a_n$, and $\limsup a_n := \inf_N a_N^+ = \inf_N \sup_{n\geq N} a_n$. I also use the standard definition of $\sup$ for a subset $E$ of the extended reals $\mathbb{R}^* = \mathbb{R} \cup{ \{\infty, -\infty \}}$ (see definition 6.2.6 in Analysis I by Terrance Tao).
Suppose $\{a_n\}$ does not have an upper bound. Then $\forall n, a_n^+ = \infty$, so $\limsup a_n = \inf \{\infty\} = \inf \{\} = \infty$; the claim immediately follows from the standard arithmetic on $\mathbb{R}^*$.
Similarly, if $\{a_n\}$ does not have a lower bound, we have $\liminf a_n =\sup \{-\infty\} = \sup \{\} = -\infty \leq \limsup a_n$.
We're now left with the case where $\{a_n\}$ is bounded, which is handled by @Pedro's answer.
