This is rather silly, I expect Asaf will point out what I am missing immediately.

Let $U$ be a filter on $\mathbb{N}$. If $\{a\_n\}\_{n=1}^\infty$ is a sequence of reals, we write $\lim\_U a\_n = a$ if for each $\epsilon > 0$ 
$$
\{ n \ : \ |a\_n - a|< \epsilon \} \in U
$$
I want to show that if $U$ is an ultrafilter, then $U$ limits exist for every bounded sequence of reals $\{a\_n\}$. Showing they are unique is easy. Also, this is easy if $U$ is principle (if it's generated by $n$, then $a\_n$ is the limit). Additionally, I have proved the result when the sequence is convergent (As, assuming $U$ is non-principle, $U$ contains all the co-finite sets).

Haven't gotten much further after playing around. I tried breaking $a\_n$ into disjoint convergent subsequences (you can do this but there may be $\omega$ many). But the lack of unity in my approach makes me think I am going about this the wrong way. Anyone have a proof?