How does the Axiom of Extensionality imply that there is exactly one empty set? In his book Notes on Set Theory, Moschovakis begins to list the axioms of $\text{ZFC}$ on p. 24. The first is the Axiom of Extensionality, which he expresses like this:
$$
A = B \Longleftrightarrow (\forall x)[x \in A \Longleftrightarrow x \in B].
$$
He then goes on to list Emptyset Axiom:

There is a special object $\emptyset$, which we will call a set, but which has no members.

He then goes on to note:

The Axiom of Extensionality implies that only one empty set exists, ...

Now, here's my problem. In its formulation, the Axiom of Extensionality uses the membership of some existing "thing" $x$ to establish "sameness" of two sets. But since $x$ is an existing "thing", how is Extensionality even relevant to $\emptyset$, which, by definition, contains no existing "things"? It seems to me that we cannot establish that there exists exactly one $\emptyset$, since the very notion of "sameness" is built upon existing $x$'s being members of sets, and $\emptyset$ contains no existent $x$'s.
Addendum
So if I understand correctly, my misinterpretation of Extensionality lies in this: I interpreted $x \in A$ as a demand for existing $x$'s to be a part of sets in order for them to be the same set. But what Extensionality really says is that:

$A = B \Longleftrightarrow (\forall x)[x \in A \Longleftrightarrow x \in B]$ holds for all $x$'s.

Then indeed, it follows that:
$$
A = \emptyset, \; B = \emptyset' \\
A = B \Longleftrightarrow (\forall x)[x \in A \Longleftrightarrow x \in B],
$$
since it holds for both $A$ and $B$ that they have no members, hence every member that is a member of $A$ will be a member of $B$ — that is to say, no member.
While thinking about it, I came up with an interesting "modification" of $\text{ZFC}$. Instead of conceiving of $\emptyset$ as a set without members, let us denote nothing as $\mathfrak{n}$. $\mathfrak{n}$ is an element of every set. More specifically, we define $\emptyset$ as $\emptyset := \{ \mathfrak{n} \}$. With this setup, my interpretation of Extensionality holds for all sets. The only drawback is that such a setup contains atoms (one atom, $\mathfrak{n}$, to be precise), which violates the Principle of Purity. Well, just a thought I wanted to share. ;)
Please correct me if I got something wrong.
 A: The notion of "sameness" is not built upon existing $x$'s being members of sets. All we care about are the implications if $x\in A$, then $x\in B$ and if $x\in B$, then $x\in A$. If $A$ and $B$ are empty sets $x\notin A$ and $x\notin B$, hence these implications are (vacuously) true. I guess the concept of vacuous truth is what you are looking for.

There is a problem with this modification: we still can construct the empty set (in the old sense) due to the axiom of specification. For instance, consider $\{x\ |\ x\in\{\mathfrak{n}\}\text{ and }x\neq x \}$. So the requirement that every set contains $\mathfrak{n}$ doesn't hold.  
A: Short answer : saying that "set $A$ = set $B$" is not a positive assertion as to the existing members of $A$ and of $B$; it is a negative statement as to the non-existence of any $x$ belonging to one set but not to the other.


*

*The temptation to over-interpret the conditional ( and consequently the biconditionnal) operator featuring in the definition of set identity must be resisted.


*In fact, the proposition " for all $x$ , $x$ is in $A$ $\rightarrow$ $x$ is in $B$ " has no existential import. Actually ( and that can be seen by inspecting the truth table of the "if...then..." operator) a conditional says very little, and what it says is purely negative: asserting "$A\rightarrow$ $B$" amounts to saying " we are not on line $2$ of the truth table" , that is, " it is not the case that $A$ is true while $B$ is false".


*So " for all x, $x$ is in $A$ $\rightarrow$ $x$ is in $B$ " simply means :
"it is not the case that there exists an $x$ such that $x$ is in $A$ while $x$ is not in $B$"***


*Therefore, the test for sameness amounts to asking 2 questions :
(1) does any existing $x$ belong to $A$ but not to $B$?
(2) does any existing $x$ belong to $B$, but not to $A$?
and a positive answer to the whole test requires 2 negative answers to these questions. ( that is, 2 assertions of non-existence).

*

*So let's say that Empty$_1$ and Empty$_2$ are two ( alledgedly distinct) empty sets.

(1) can you see an existing $x$ that belongs to Empty$_1$ but not to Empty$_2$ ?
and
(2) can you see an existing $x$ that belongs to Empty$_2$ but not to Empty$_1$?

*

*In case you have answered "no" to these 2 questions, you know that these two empty sets are, in fact, one and the same set.

