In Halmos' Naive set theory, on p. 35, he writes
An empty union makes sense (and is empty), but an empty intersection does not make sense.
Suppose my index set is $I = \{1, 2\}$ and my indexed sets are defined as $A_i \equiv \{-i, i\}$. Then $A_1 = \{-1,1\}$, $A_2 = \{-2, 2\}$, $\bigcup A_i = \{-2, -1, 1, 2\}$ and $\bigcap A_i = \emptyset$. I don't see why that doesn't "make sense".
Consider a family of sets $A_i$ indexed by $i \in I$. What is the definition of $\bigcap\limits_{i \in I} A_i$? It is supposed to be $\{x : \forall i \in I, x \in A_i\}$.
In other words, it's the set of all $x$ which are in all of the $A_i$.
Now, we must ask ourselves whether this intersection in fact forms a set.
Theorem: $\bigcap\limits_{i \in I} A_i$ forms a set iff $I$ is nonempty.
Proof: suppose that $I$ is non-empty. That is, we have some $j \in I$. Then we note that $\{x : \forall i \in I, x \in A_i\} = \{x \in A_j : \forall i \in I, x \in A_i\}$. And the latter is always a set.
Now suppose $I$ is empty. Then $\{x : \forall i \in I, x \in A_i\}$ is the set of all sets. But it's well-known that there can be no set of all sets. This is a contradiction. So $\bigcap\limits_{i \in I} A_i$ does not exist.
OP's confusion arises from the term "empty intersection". The author means that the indexing set is empty, not that the resulting intersection is empty.
