How do you tackle an infinite set when proving some property in set theory? There is a property in set theory:

$B\cup \bigcap_{\lambda\in V}A_{\lambda}=\bigcap_{\lambda\in V}(B\cup A_{\lambda})$

And as well as other familiar properties, the set $V$ can be chosen arbitrarily, including an infinite set. I know the property can proved by induction since $A\cup(B_1\cap B_2)=(A\cup B_1)\cap(A\cup B_2)$ is true. But how can I prove it when the set $V$ is infinite?
Thank you for any help or idea!
 A: You don't need induction. Just show two sided inclusion. For example, let's how the inclusion $\subseteq$. Assume $x\in B\cup (\cap_{\lambda\in V} A_{\lambda})$. We want to prove that $x\in \cap_{\lambda\in V} (B\cup A_{\lambda})$. So let $\lambda_0\in V$. By assumption we have $x\in B$ or $x\in\cap_{\lambda\in V} A_{\lambda}$. If $x\in B$ then $x\in B\cup A_{\lambda_0}$. If $x\in\cap_{\lambda\in V} A_{\lambda}$ then in particular $x\in A_{\lambda_0}$ and thus $x\in B\cup A_{\lambda_0}$. So in either case we proved that $x\in B\cup A_{\lambda_0}$. Since this is true for all $\lambda_0\in V$ this implies $x\in\cap_{\lambda\in V} (B\cup A_{\lambda})$.
Now, the other inclusion is similar.
A: By definition $x \in \bigcap\limits_{\lambda\in V}A_{\lambda} \Leftrightarrow \forall \lambda\in V, x \in A_{\lambda}$ so using this
$$x \in B \cup \bigcap\limits_{\lambda\in V}A_{\lambda} \Leftrightarrow \forall \lambda\in V, x \in B \lor  x \in A_{\lambda}\Leftrightarrow x \in \bigcap\limits_{\lambda\in V}A_{\lambda} \cup B$$
