Let $\omega$ be the set of all finite ordinals, i.e. $\omega=\{0,1,2,...\}$ where $0=\emptyset$ $1=\{0\}=\{\emptyset\}$ $2=\{0,1\}=\{\emptyset, \{\emptyset\}\}$ etc...
A set $A$ is called an inductive set if for each $n\in A$, $n+1=n\cup\{n\}\in A$.
So the problem is, am I allowed to use proof by induction to prove that $\omega=\mathbb{N}$ is inductive?
If not, what should I do?
You don't need induction for this proof, granted you have some theorems under your belt.
Specifically, you want that the union of two finite sets is finite, which is generally proved using induction.
Once you have that, you can show that $n$ is a finite ordinal if and only if $n\cup\{n\}$ is a finite ordinal. That is pretty much immediate from the definition.
Now, if $n$ is finite, and $\{n\}$ is certainly finite, then $n\cup\{n\}$ is also finite. Therefore $n\cup\{n\}$ is a finite ordinal. So $n\in\omega$ if and only if $n\cup\{n\}\in\omega$ as well.
So the induction was only used to prove that $n\cup\{n\}$ is finite, assuming that $n$ is finite.