the intersection of inductive sets is a set In these notes 
http://home.fau.edu/wmcgove1/web/Courses/Ntransitive.pdf
Definition 1.3 it is said that the intersection of inductive sets is itself inductive.
A set $A$ is inductive if: $(\emptyset \in A) \land (n \in A \implies n^+ := n \cup \{n\} \in A)$.
I think we cannot say that the class of all inductive sets is itself a set (can we?). 
So what we really want to prove is that if $A$ and $B$ are inductive sets then $(A\subset B) \lor (B\subset A)$ so that if an inductive set $A$ actually exists (this is eventually guaranteed by an appropriate axiom) then every inductive set either contains $A$ or is contained in $A$ and the intersection of all inductive subset of $A$ is itself inductive and hence is the smallest inductive set.
If this is correct the question is: how do we prove that either $A\subset B$ or $B\subset A$?
 A: You cannot prove that every two inductive sets satisfy $A\subseteq B$ or $B\subseteq A$.
What you can prove, however, is that the intersection of any number of inductive sets is an inductive set. From this we can conclude that if $B$ is an inductive set, then there is a smallest inductive set $A$ such that $A\subseteq B$: simply take the intersection of all the inductive subsets of $B$.
But now I claim even more, this $A$ is in fact a subset of any inductive set. To see why, simply note that if $C$ is an inductive set, then $B\cap C$ is an inductive subset of $B$, and therefore $A\subseteq B\cap C$. This $A$ is in fact $\omega$, the least infinite ordinal.
Now, if $\alpha$ is a limit ordinal, then $\alpha$ is an inductive set. But also $V_\alpha$, the $\alpha$th stage of the von Neumann hierarchy is an inductive set. Take any limit ordinals $\alpha$ and $\beta$ such that $|V_\alpha|<|\beta|$, and you have two inductive sets which are not subsets of one another. (And it follows also that the class of inductive sets is not a set, indeed.)
