Other's have given fantastic analogies and examples, but I haven't seen anyone formally go through why the empty set is unique.
Given that we have the Axiom of Set Existence and the Comprehension Axiom Schema
The existence of the empty set is given by the following theorem.
Theorem
$\exists$y$\forall$x(x$\in$y $\iff$ x$ \neq$ x)
Proof:
By the Set Existence Axiom: $\exists$x$(x=x)$, fix $x_0$ s.t. $x_0$ = $x_0$
Consider $\phi$(z) := z $\neq$ z
Then, By the Axiom Schema of Comprehension
$\exists$y$\forall$z(z$\in$y $\iff$ $\phi$(z) $\wedge$ z$\in$$x_0$)
Thus,
$\exists$y$\forall$z(z$\in$y $\iff$ z $\neq$ z $\wedge$ z$\in$$x_0$)
Equivalently, $\exists$y$\forall$z(z$\in$y $\iff$ z $\neq$ z)
As z $\neq$ z $\iff$ z $\neq$ z $\wedge$ z$\in$$x_0$
Informally,
y = { z$\in$$x_0$ | z $\neq$z} = {z | z $\neq$ z} = {z| $\phi$(z) } exists ( is a set)
Why is such a y unique?
Informally, uniqueness means that if $y_1$ = {w| $\phi$(w) } and $y_2$ = {z| $\phi$(z) }, then $y_1$ = $y_2$ i.e. rather than " a empty set". We can refer to it as "the empty set".
We need the Axiom of Extentionality.
Theorem:
$\exists!$y$\forall$x(x$\in$y $\iff$ x$ \neq$ x)
i.e.
$\exists$$y_1$$[\forall$x(x$\in$$y_1$ $\iff$ x$ \neq$ x) $\wedge$ ($\forall$$y_2$$\forall$x(x$\in$$y_2$ $\iff$ x$ \neq$ x) $\rightarrow$ $y_1$ = $y_2$)]
Proof:
Fix $y_1$,$y_2$ s.t.
$\forall$x(x$\in$$y_1$ $\iff$ x$ \neq$ x)
$\forall$x(x$\in$$y_2$ $\iff$ x$ \neq$ x)
We prove that $y_1$ = $y_2$
$\forall$x(x$\in$$y_1$ $\iff$ x$ \neq$ x $\iff$ x$\in$$y_2$)
and so $\forall$x(x$\in$$y_1$ $\iff$ x$\in$$y_2$), thus by the Axiom of Extentionality $y_1$ = $y_2$
We have verified the existence and uniqueness of the empty set.
With this Theorem $\exists!$y$\forall$x(x$\in$y $\iff$ x$ \neq$ x), we can unambiguously denote the empty set with the defined term:
$\emptyset$ = The unique y s.t. $\forall$x(x$\in$y $\iff$ x$ \neq$ x)