Let $(a_n)$ be a sequence of real numbers which is bounded above and $a_n\not\rightarrow -\infty$ then $a_n$ has a convergent subsequence.
I think I will be done if I can show that there is a $M\in\mathbb{R}$ such that for each $n_0$ there is an $n>n_0$ such that $a_n\ge M$. Please help.
If you are familiar with the definition of $\limsup$, this answers your question.
By hypothesis $\exists M\in\mathbb{R}$ s.t. $a_n\leq M,\ \forall n\in\mathbb{N}$.
Set $b_m$:=$sup_{k\geq m}a_k$.
$\{b_m\}_{m\in\mathbb{N}}$ is a decreasing sequence and ${b_m}\leq M$. Moreover $b_m$ is a convergent sequence and $L:=\lim_{m\rightarrow\infty}b_m=\limsup_{m\rightarrow\infty}a_m>-\infty$, otherwise by the sandwich theorem you would have $\lim_{n\rightarrow\infty}a_n=-\infty$. Now you are done since it is known (otherwise) that there exists a subsequence converging to the $\limsup$.
In general, you cannot bound $a_n$ from below since you can have something like $$a_n=
\left\{ \begin{array}
- -n\ \ \ if\ n \ is\ not\ a\ multiple\ of\ 10\\
1\ \ \ \ \ if\ n\ is\ a\ multiple\ of\ 10
\end{array}\right.$$
