The intersection of $3$ set is empty, would the intersection of $4$ sets be empty? Let me clarify some more.
Let's say we have four sets $A,B,C,$ and $D$.
If the intersection of any three sets is empty, by default is the intersection of all four sets empty?

Yes, if $A$, $B$, $C$, and $D$ are sets such that the intersection of any three is empty, then because intersection is associative, we have that $A \cap B \cap C \cap D = (A \cap B \cap C) \cap D = \emptyset \cap D = \emptyset$.
Just in case you aren't sure what associative means, it means that we can intersect in any order.  So, in the case of three sets, we get $(A \cap B) \cap C = A \cap (B \cap C)$.  In the case of four sets intersected, you can intersect them in any order you want.
