Any subset of elements in an ultrafilter is in the ultrafilter I am trying to prove the following fact:

Given $\mathcal{F}$ an ultrafilter on $X$, suppose $ A \in \mathcal{F}$ and $B \subseteq A$, then  either $B$ or $A\backslash B$ are in $\mathcal{F}$

Recall:

$\mathcal{F}$ is an ultrafilter if for all $F$ a filter on $X$,
  $\mathcal{F} \not\subset F$

My approach: 
Let $A \in \mathcal{F}$, and $B \subseteq A$, then suppose that $B \not \in \mathcal{F}$, we will show that $A \backslash B$ is in $\mathcal{F}$. 
How to proceed from here?

You already know that either $B$ or $X\setminus B$ is in $\mathcal F$.
Therefore if $B\notin\mathcal F$ it must be that $X\setminus B\in\mathcal F$, and since a filter is closed under finite intersections, we also have
$$ A \cap (X\setminus B) \in \mathcal F $$
