Empty set, subsets, and vacuous truths So I was trying to prove that a null set is a subset of any set.
First, to define when $A$ is a subset of $B$:
$$A \subseteq B \iff \forall x(x \in A \implies x \in B)$$
(At least I think that's right?)
So then consider the empty set $\emptyset$ for which $\forall x(x \not\in \emptyset)$ is true.
I tried to prove that the empty set is a subset of any other set, or:
$$\emptyset \subseteq B \iff \forall x(x \in \emptyset \implies x \in B) \vdash \text{T}$$
To me this seemed true because it was "vacuously true"... somehow. Like it makes sense to call it vacuously true that "all $0$ items in $\emptyset$ can be found in $B$, yep!" but that isn't satisfying to me, how do I "prove" this is the case? 
Is $x \in \emptyset$  a false... statement? A false predicate? Something else? Something that results in false so that the implication itself is true. Is this a $(\text{F}\implies \text{F}) \vdash \text{T}$ thing?
Or is it the $\forall$ that makes it false somehow?
What if I had said $\exists x \in \emptyset$, this feels like it would certainly be false but again I can't prove why, it's just an intuition.
Can anyone clarify the correct definitions / implications and why they're true or false or what have you? 

The proof relies on Ex falso :

$\vdash \lnot P \to (P \to Q)$.

We have to apply it in the form :

$\lnot (x \in \emptyset) \to (x \in \emptyset \to x \in B)$.

We have (axiom or theorem) : $\mathsf {ZF} \vdash \forall x \ \lnot (x \in \emptyset)$.
By Universal instantiation we get : $\lnot (x \in \emptyset)$ and thus from Ex falso, by Modus Ponens : $(x \in \emptyset \to x \in B)$.
Finally, by Universal generalization we conclude with :

$\mathsf {ZF} \vdash \forall x \ (x \in \emptyset \to x \in B)$.

