Proving intersection of empty set is the set of all sets

Question

In an exercise, it asks to prove $\bigcap \varnothing$ is equal to the set of all sets. I understand that there are many proofs online but I have a specific question (mostly regarding mathematical logic) regarding my proof.

From my understanding, given a set $S$ (where all of its elements are also sets), $x \in \bigcap S \iff \forall \, Y\in S, x \in Y$.

Now I will try and show the set of all sets is contained in $\bigcap \varnothing$. Let $x$ be an arbitrary set - now $x \in \bigcap \varnothing \iff \forall \, Y\in \varnothing, x \in Y$. My question is why the statement $\forall \, Y\in \varnothing, x \in Y$ vacuously true? From my understanding, a statement is vacuously true in the case of an implication (ie. $p \implies q$ where $p$ is false) but I fail to see how $\forall \, Y\in \varnothing, x \in Y$ can be written in an implication (I tried writing it in the form $\forall Y: Y \in \varnothing \implies x \in Y$ but this is not the same statement)

Answer 1

$\forall Y \in \varnothing \; P(Y)$ means $\forall Y \; (Y \in \varnothing \implies P(Y))$. But $Y \in \varnothing$ is always false, making the implication $Y \in \varnothing \implies P(Y)$ true.

Answer 2

Since $\emptyset$ has no elements, each unary predicate $\varphi$ satisfies $\forall Y\in\emptyset(\varphi(Y))$. In this case, our predicate is $x\in Y$.

Answer 3

Consider the negation of $\forall Y\in\varnothing, x\in Y$. It is $\exists Y\in\varnothing, x\notin Y$. The latter is false, because there is no element in the empty set. Hence the original proposition was true.