Proof: $U$ is an ultrafilter of a Boolean algebra $B$ if and only if for all $x$ in $U$ exactly one of $x$,$x^*$ belongs to $U$.

I have been stuck with this problem for a while now. I have a proof that letting $ U$ be an ultrafilter, exactly one of $ x,x^*$ belongs to $ U$ for all $ x$ in $ B$ , I did this by showing that both belong to $ U$ implies that $ U=B$ which cannot be the case, and if neither long to $ U$ then a contradiction can be derived by de-Morgan’s laws. However, I’m stuck with the reverse implication, I need to prove that:

“If $ U$ is a subset of a Boolean algebra $ B$ such that $ \forall x \in B$ , exactly one of $ x \in U$ and $ x^* \in U$ is true.

Is this even the case? Any help would be greatly appreciated, thanks in advance.