Distributive law for indexed sets.? I want to prove .
$A\cap(\bigcup_{\lambda \in \Lambda} B_{\lambda})=\bigcup_{\lambda \in \Lambda} (A\cap B_{\lambda})$
My proof:
Let $x\in  A\cap(\bigcup_{\lambda \in \Lambda} B_{\lambda})$
$\implies x\in A  \text{ and } x\in \bigcup_{\lambda \in \Lambda} B_{\lambda}$
$\implies x\in A \text{ and } x \in B_{\lambda} \text{ for all } \lambda \in \Lambda $
$\implies x\in A\cap B_{\lambda} \text{ for all } \lambda \in \Lambda .$
Then $x \in \bigcup_{\lambda \in \Lambda}A \cap B_{\lambda} $. The backward direction is similar. I feel that the proof is not correct.
 A: It's quite okay, except the fact that $x\in B_\lambda$ for all $\lambda$ is wrong. It should be for some $\lambda \in \Lambda$. All the consequent "for all"s will also be "for some".
This is because you have $x\in \cup_{\lambda \in \Lambda}B_\lambda$ which means there is a $\lambda_0 \in \Lambda$ such that $x\in B_{\lambda_0}$
A: $\newcommand{\l}{\lambda}
\newcommand{\ll}{{\lambda_0}}
\newcommand{\b}{\bigcup_{\l \in \Lambda} B_\l}
$
One error in your argument is this:
$$x \in \bigcup_{\l \in \Lambda} B_\l \implies (\exists \l \in \Lambda)(x \in B_\l)$$
That is, if $x$ is in the union, you can only ensure that it is in at least one of the $B_\l$, not all of them.
If we call this $\l$ by $\l_0$ instead, then we see that
$$x \in B_{\l_0} \land x \in A \iff x \in A \cap B_{\l_0}$$
Then, taking the union over all $\l \in \Lambda$ gives the desired result.

Really, this proof can be easily reversed as well. It is noteworthy that if you want to write a chain of implications as you did, e.g.
\begin{align*}
P \iff& Q \\
\implies& R
\end{align*}
and such, you shouldn't just focus on writing the implications in one way. For instance, I wouldn't write
$$\begin{align*}
x \in A \cap B
\implies& x \in A \land x \in B \\
\implies& x \in A
\end{align*}$$
but rather make the first arrow a biconditional $\iff$. Technically, yes, if your goal is just to show one direction, it doesn't matter. But especially if you have a proof where you need both directions, and you can connect both "ends" with biconditionals, then you can write your proof neatly and compactly.

As an example, this is how I would write the correct proof for this claim:
$$\begin{align*}
x \in A \cap \b
\iff& x \in A \land x \in \b  \\
\iff& x \in A \land (\exists \ll \in \Lambda)(x \in B_\ll) \\
\iff& (\exists \ll \in \Lambda)(x \in A \land x \in B_\ll)  \\
\iff& (\exists \ll \in \Lambda)(x \in A \cap B_\ll)  \\
\iff&  x \in \bigcup_{\l \in \Lambda} (A \cap B_\l)
\end{align*}$$
