The union of a set with a set that has the finite intersection property has F.I.P 
Let $I$ be a set and let $C$ be a collection of subsets of $I$ with the finite intersection
property. Prove that for every $A \subset I$, either $C \cup \{A\}$ or $C \cup \{A^{c}\}$ has the finite intersection property.

I am trying to prove the above statement and I am just not sure where to even start. I know that since C has the finite intersection property that the intersection of any finite subcollection of C is nonempty.
Do i need to look at different cases? If $A \in C$ and $A \notin C$.
If $A \in C$ does that automatically mean $C \cup \{A\}$ has F.I.P. or no?
And if $A \notin C$ does that imply $A^c \in C$ since $A \subset I$?
 A: I’ll use $\mathscr{C}$ instead of $C$ for the family with the finite intersection property.
Yes, if $A\in\mathscr{C}$, then $\mathscr{C}\cup\{A\}=\mathscr{C}$, so it has the finite intersection property. However, $A\notin\mathscr{C}$ does not imply that $I\setminus A\in\mathscr{C}$. Suppose, for instance, that $A_n=\{k\in\Bbb N:k\ge n\}$ for each $n\in\Bbb N$, and $\mathscr{C}=\{A_n:n\in\Bbb N\}$; it’s easy to check that $\mathscr{C}$ has the finite intersection property. Let $A=\{2n:n\in\Bbb N\}$, the set of even natural numbers, so that $\Bbb N\setminus A$ is the set of odd natural numbers; then neither $A$ nor $\Bbb N\setminus A$ belongs to $\mathscr{C}$.
HINT: I suggest trying to prove the contrapositive: show that if neither $\{A\}\cup\mathscr{C}$ nor $\{I\setminus A\}\cup\mathscr{C}$ has the finite intersection property, then $\mathscr{C}$ does not have the finite intersection property. Start by noticing that if $\{A\}\cup\mathscr{C}$ does not have the finite intersection property, then there is a finite $\mathscr{C}_0\subseteq\mathscr{C}$ such that $A\cap\bigcap\mathscr{C}_0=\varnothing$. There’s a further hint in the spoiler box if you need it.

 If $A\cap X=\varnothing=B\cap X$, then $$(A\cup B)\cap X=(A\cap X)\cup(B\cap X)=\varnothing\,.$$

