For two collections of sets $\{A_i\}_{i\in\mathbb{N}}$ and $\{B_i\}_{i\in\mathbb{N}}$ the question is to prove that $$\bigcup\limits_{i\in\mathbb{N}}A_i\backslash\bigcup\limits_{i\in\mathbb{N}}B_i\subset\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]$$ My attempt: $$\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]=(\bigcup\limits_{i\in\mathbb{N}}[A_i\cap\{\bigcup\limits_{n\in\mathbb{N},\ n\ne i}B_n\}])\ \cup\ [\bigcup\limits_{i\in\mathbb{N}}A_i\backslash\bigcup\limits_{i\in\mathbb{N}}B_i]$$ So $\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]$ is the union of two things, one of which is the thing we want prove that it is included in $\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]$
Can sombody confirm or disprove what I did please?
HINT
In such cases it is often useful to work with elements of the relevant sets. Show that for any $x\in\bigcup A_i\setminus \bigcup B_i$ holds $x\in\bigcup (A_i\setminus B_i)$.
Note that $x\in A\setminus B \Leftrightarrow x\in A \wedge x\notin B$ and $$x\in\bigcup_{i\in\mathbb{N}} C_i \Leftrightarrow \exists\; k\in\mathbb{N}:x\in C_k$$
EDIT: On popular demand, I will show why
$$\bigcup\limits_{i\in\mathbb{N}}[A_i\backslash B_i]=(\bigcup\limits_{i\in\mathbb{N}}[A_i\cap\{\bigcup\limits_{n\in\mathbb{N},\ n\ne i}B_n\}])\ \cup\ [\bigcup\limits_{i\in\mathbb{N}}A_i\backslash\bigcup\limits_{i\in\mathbb{N}}B_i]$$
does not hold. (For me, $\mathbb{N}=\{1,2,3,\ldots\}$. Not relevant for the proof, just noticed it to avoid confusion.)
Set $A_1:=\{0\}$, $A_2=A_3=A_4=\ldots:=\emptyset$ and $B_1=B_2=B_3=\ldots:=\{0\}$. Clearly $\bigcup (A_i\setminus B_i)=\emptyset$ and $\bigcup A_i \setminus \bigcup B_i=\emptyset$ but $$\bigcup_{n\in\mathbb{N},n\neq i}B_n=\{0\}$$ for all $i\in\mathbb{N}$, therefore $$A_1\cap\{\bigcup\limits_{n\in\mathbb{N},\ n\ne 1}B_n\}=\{0\}$$ and so the right hand side of the equation is $\{0\}$ while the LHS is $\emptyset$, contradiction.