Here we extend our language by the symbol $\emptyset$ (sometimes $\varnothing$ or $\{\}$ is used also), denoting the term (set) with the defining property

$\emptyset=\{x\mid \bot\}$

We're thus claiming

$\exists y.\,y=\{x\mid \bot\}$

which is short for

$\exists y.\,\forall x.\,\left(x\in y\Leftrightarrow \bot\right)$

which is short for

$\exists y.\,\forall x.\,\left((x\in y\implies \bot)\land(\bot\implies x\in y)\right)$

Is this true? Does our set theory permit the existence of such a set $y$? Yes, it's granted by the axiom (axiom of empty set (Wikipedia))

$\exists y.\,\nexists x.\,x\in y$

(in words: “There exists a set $y$, such that there does not exists an $x$ which is in it.”)

which is equivalent to

$\exists y.\,\forall x.\,\neg(x\notin y)$

i.e.

$\exists y.\,\forall x.\,\left(x\in y\implies\bot\right)$

As $\bot\implies P$ is $\top$ for any $P$ and as $Q\land\top$ is logically equivalent to $Q$, we are done.

One can show that there is at most one such set (from the extensionality axiom of set theory, and if the domain of discourse is non-empty),

$\exists! y.\,y=\{x\mid \bot\}$

i.e.

$\exists! y.\,\forall x.\,\left(x\in y\implies\bot\right)$

which is short for

$\exists y.\ \forall z.\ (\left(\forall x.\,\left(x\in z\implies\bot\right)\right) \Leftrightarrow y=z)$

Remark: We could also avoid the symbol $\bot$ and use the predicate $x\neq x$.

Wikipedia: Empty set, Axiom of empty set (Wikipedia)

First infinite von Neumann ordinal