This kind of result is often most easily proved by ‘element-chasing’: let $x$ be any element of the set on the lefthand side, and show that $x$ must be an element of the set on the righthand side. (If you were proving an equality, instead of an inclusion, you’d then repeat the process in the other direction.)
So suppose that $x\in(A_1\cap A_2\cap\ldots\cap A_n)\triangle(B_1\cap B_2\cap\ldots\cap B_n)$. Then either $$x\in A_1\cap A_2\cap\ldots\cap A_n\quad\text{and}\quad x\notin B_1\cap B_2\cap\ldots\cap B_n\;,\tag{1}$$ or $$x\in B_1\cap B_2\cap\ldots\cap B_n\quad\text{and}\quad x\notin A_1\cap A_2\cap\ldots\cap A_n\;.\tag{2}$$ The two cases are entirely similar, so let’s assume $(1)$.
Since $x\notin B_1\cap B_2\cap\ldots\cap B_n$, there is some $k\in\{1,\dots,n\}$ such that $x\notin B_k$. But $x\in A_k$, so ... ?
Your proof of the one direction looks perfectly fine. It is very common to use "and" and "or" written in a meta-level proof. However, if you feel this getting into the way of displaying an argument, you may benefit by some more formalism, e.g.
\begin{align}x\in A\cap(B\cup C) &\Rightarrow x\in A\land x\in (B\cup C)\\ &\Rightarrow x\in A\land (x\in B\lor x\in C)\\&\Rightarrow (x\in A\land x\in B)\lor (x\in A\land x\in C)\\&\Rightarrow x\in A\cap B\lor x\in A\cap C\\ &\Rightarrow x\in(A\cap B)\cup (A\cap C)\end{align}
I'm using $\land,\lor$ here as symbols of the "and","or" respectively.
I this specific case, the proof relies on using the distributive property of $\land,\lor$ as logical connectives to prove the corresponding property of unions and intersections via set membership. (this connection of course comes as no surprise through the common connection to boolean algebra(s)).
So, having this translation of very similar connectives/operations into one another as the essence of a proof can seems a little ambiguous, although it is the heart of the argument.
In general, I think the key to a clear exposition of course always depends on the context, but it is almost always a good mix of formatting, formalism and non-formalism.
E.g. no one wants to write precise first-order set theory statements for every mathematical concept, although it would be formally strict. In cases like the above, or in general with chains of implications or bi-implications, a structured formatting of the steps, like the aligned presentation above, can already improve the exposition a lot.
EDIT: Note, that you can turn every implication in the above chain into a bi-implication, i.e. by the argument above, you may prove directly that $x\in A\cap (B\cup C)\Leftrightarrow x\in(A\cap B)\cup (A\cap C)$. Thus, by extensionality of sets, you straightforwardly have $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$.
EDIT2: The distributive property of the logical connectives $\land,\lor$ may be verified by corresponding truth tables. It is important to note, that $\cap,\cup$ are defined operations in the theory of sets while the underlying logic(where you proceed with your reasoning with $\land,\lor$) is the first-order logic(of set theory). The distributive property of the logical connectives is a theorem of first-order logic which can then be used in your proof to apply it to propositions about the set-membership relation.
The reasoning is less circular as it is referential. In the end, you derive a property of your defined connectives by using a similar property of the connectives used inside their formal definition.
Best Answer
You have $$( B_1\cap B_2)\cup C = (B_1\cup C)\cap(B_2 \cup C).$$ Use an induction argument.