Is the intersection of every non-empty family of inductive sets equal to the intersection of every inductive set? In Halmos Naive set theory, there is the following passage (excuse my french) in his section introducing natural numbers : 

In this language the axiom of infinity simply says that there exists a successor [inductive] set A. Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?), the intersection of all the successor sets included in A is a successor set $\omega$.

I have trouble seeing why one should feel the need to state the bolded part in order to derive the existence of $\omega$.
Would he not have arrived at the same conclusion had he decided to consider directly the intersection of every inductive set included in A? 
Would he not also have done so had he decided to state that the intersection of every inductive set was also an inductive set instead of invoking families of inductive sets?
 A: The intersection of all happy sets is always a good candidate for the smallest happy set. However, you do need to show that such an intersection is still happy (otherewise ist just a rather small set, but not the smallest happy set). Also, we run into some logical problems if we want to take the intersection of all happy sets as the happy sets may form a proper class! This can be circumvented by grabbing an arbitrary happy set, taking the intersection of all its happy subsets and showing that the same intersection is obtained if one starts with any other happy set. All this follows if one has that the intersection of any nonempty family of happy sets is happy -  and that there exists at least one happy set.
With happy=inductive, you have the situation of your question: The existence of at least one inductive set is the Axiom of Infinity; the intersection property is the bold part questioned by you.
A: Note that the definition of inductive is this:


*

*$\varnothing\in A$;

*$\forall x(x\in A\rightarrow x\cup\{x\}\in A)$.


The first property is definitely preserved by intersections. To see that the second property holds for intersections, if $A_i$ is a set of inductive sets then $x\in A_i$ for all $i$ means that $x\cup\{x\}\in A_i$ for all $i$. Therefore both properties must hold when taking intersections.
Generally speaking one can replace the second condition by any $\forall x\exists y(\ldots)$ where $\ldots$ define $y$ to be some unique set (e.g. $y=x\cup\{x\}$).
