How can I define $\mathbb{N}$ if I postulate existence of a Dedekind-infinite set rather than existence of an inductive set? Suppose in the axioms of $\sf ZF$ we replaced the Axiom of infinity 

There exists an inductive set. 

with the Axiom of Dedekind-infinite set

There exists a set equipollent with its proper subset.


How can I define the set of natural numbers $\mathbb{N}$ in this setting, and prove that it is the unique minimal inductive set?

Suppose that $A$ is a Dedekind-infinite set. First consider $T=\operatorname{TC}(A)$, the transitive closure of $A$. Now consider the function $f(x)=\operatorname{rank}(x)$, whose domain is $T$.
By the axiom of replacement the range of $f$ is a set, and it is not hard to prove that it has to be an ordinal.
Finally, prove by induction that if $n$ is a finite ordinal,1 then there are no Dedekind-infinite sets of rank $n$ (we don't need an inductive set, if such set doesn't exist then this is just an induction on the class of ordinals). And therefore there is an infinite ordinal in the range of $f$. Take $\omega$ as the least such ordinal.


*

*It is easy to define a finite ordinal if you already know what $\omega$ is, but in its absence you can define a finite ordinal to be a Dedekind-finite ordinal; or if you really like then you can use one of the many other formulations of finiteness. My favorite is due to Tarski:

$A$ is finite if and only if for every $U\subseteq\mathcal P(A)$ which is non-empty, there is a $\subseteq$-maximal element in $U$.

