Munkres suggests that we can definite the infinite intersection of an empty collection $\mathcal{A}$ of sets as $$\bigcap\limits_{A \in \mathcal{A}} A = \mathcal{U},$$ where $\mathcal{U}$ is the entire set-theoretic universe. This means sense intuitively, since every $x \in \mathcal{U}$ trivially lives in every $A \in \mathcal{A}$.
On the flip side, for some set $B$, we consider $$\bigcap\limits_{A \in \mathcal{A}} (B-A),$$ where $\mathcal{A}$ is still the empty collection. I am trying to frame the argument purely in terms of set differences rather than compliments, using the notation in Munkres. If I didn't do that, I would write $$\bigcap\limits_{A \in \mathcal{A}} (B - A) = \bigcap\limits_{A \in \mathcal{A}} B \cap A^c,$$ but $A^c$ doesn't make much sense since $\mathcal{A}$ is the empty collection. If $A$ lives in $\mathcal{A}$, isn't $A$ the empty set? The complement of the empty set is $\mathcal{U}$, the whole universe, and so $B \cap A^c = B \cap \mathcal{U} = B$. The intersection of $B$ with itself an arbitrary number of times is $B$. This is the result I was looking for, but I don't think the process is completely correct.
x in $\cap$empty collection iff
for all K in empty collection x in K iff
True.
Explanation of the last step.
If K in empty collection, then the contradiction empty collection is not empty. Thus False has been derived. Where upon prove x in K by contradiction.
If x not in K, then False as shown above.
By contradiction, conclude x in K.
Such is the wonderous mysteries of material implication, that False implies True.
Your set difference calculations are superfluous because in both cases the collection is empty.