Well Ordering Principle to prove the Distributive Law of union over the intersection of n sets I am looking for hints on solving the below problem from a textbook (copied as given):

Union distributes over the intersection of two sets: $A\cup (B\cap C)$
Use the above and the Well Ordering Principle to prove the
Distributive Law of union over the intersection of n sets:
$A\cup (B_1 \cap B_2 \cap B_3 ..... \cap B_{n-1} \cap B_n) = (A \cup B_1) \cap (A \cup B_2)..... \cap (A \cup B_{n-1}) \cap (A \cup B_{n})$
Extending formulas to an arbitrary number of terms is a common (if
mundane) application of the WOP.

I am fairly OK with WOP proofs as such to prove claims about non negative integers, but not sure about how to approach sets.
Proof:
Proof by WOP.
Let us assume, that the Distributive Law of union over the intersection of n sets does not hold (for getting a contradiction) for some set of sets ${A_1, B_1, A_2, B_2,...}$
Now, stuck!
WOP applies to set of positive integers, what if we have a set of sets?
 A: We may need WOP to prove this, instead we can do it by induction on $n$ (induction on natural numbers is proved by ZF to be valid):
For $n=1$, we can easily verify that $A\cup(B\cap C)=(A\cup B)\cap (A\cup C)$ by extensionality (this argument can be visualized using a Venn diagram.)
For $n>1$, assume that $A\cup(B_1\cap B_2\cap\ldots B_{n-1})=(A\cup B_1)\cap(A\cup B_2)\cap\ldots(A\cup B_{n-1})$ has been proven. Then we consider $A\cup(B_1\cap B_2\cap\ldots B_{n-1}\cap B_n)$: for clarity let $X$ be the set $B_1\cap B_2\cap\ldots B_{n-1}$. Then we have $A\cup (X\cap B_n)$, which by the $n=1$ case is equal to $(A\cup X)\cap (A\cup B_n)$, namely $(A\cup (B_1\cap\ldots B_{n-1}))\cap (A\cup B_n)$. Applying the induction hypothesis we find this equals $(A\cup B_1)\cap(A\cup B_2)\cap\ldots(A\cup B_{n-1})\cap (A\cup B_n)$, and since intersections commute we are done.

It's good to prove this without using WOP, since to use WOP in a proof we would first need to assume choice. On the other hand the induction proof works even if we don't assume choice.
