I first want to proof that the intersection of two inductively defined sets is again inductive. We have $A$ which is a set of elementary elements and a set of functions $F$ where each function has arity $k \geq 1$. The set $S$ is the smallest set inductively defined through $A$ and $F$ for which the follwing is true:
(1) $A \subseteq S$
(2) $f \in F$ and $f$ has arity $k$ and $s_1,...,s_k$ are elements of $S$. Then $f(s_1,...,s_k)$ is also in $S$.
Assume that $S_1$ and $S_2$ are both sets for which (1) and (2) are true. I want to show that the rules (1) and (2) are also true for the intersection $S_1 \cap S_2 = S$.
I am not even sure where to start. What does it even mean that $S_1$ and $S_2$ are sets for which both (1) and (2) are true? Do I know that $A_1 \subseteq S_1$ and $A_2 \subseteq S_2$ or do I know that $A \subseteq S_1$ and $A \subseteq S_2$ or is it irrelevant in this situation?
Let's assume that $A \subseteq S_1$ and $A \subseteq S_2$. Therefore we also know that $A \subseteq S$ by definition of the intersection. So (1) is true for $S$.
Is this even correct until now? I think if I wanted to show that (2) is true for $S$, I would take an arbitrary function $f$ with arity $k \geq 1$ and $s_1,...,s_k$ elements and show that $f(s_1,...,s_k) \in S$.
Intuitively, you can view $A$ as the set of "building blocks" that you begin with, and $F$ the operations that let you construct new blocks from these.
Example: let $A=\{a,b\}$ — the elements we start with — and $f: x,y \mapsto xy$, e.g. $f(a,b) = ab$, $f(aba,bb)= ababb$ and so forth. Then the set $S$ inductively generated from $A$ and $f$ in this case is the set of all strings that you can form starting from $a$ and $b$ by applying $f$ zero or more times, e.g. $ab;bbbb;abab;babaa$ are all in $S$.
I took it that you needed an informal explanation, you can see how to proceed from here? — it should be easy I guess. BTW, no requirement of unique readability, right?