Proof about Sequence of Sets Problem: Let $ A_1,A_2,...,A_n$ be sets such that $X=\bigcup_{i=1}^n A_i$.
 Prove that there exists a sequence of sets $ B_1,B_2,...,B_n$ such that
a) $B_i \subseteq A_i$ for each $i=1,2,...,n$
b) $B_i \cap B_j = \emptyset$ for $i \neq j$
c) $X=\bigcup_{i=1}^n B_i$
My observation: I look at the "new things" added to $X$ by $A_i$ and call that $B_i$. For example $B_1$ is $A_1$, $B_2$ is $A_2\setminus A_1$ and $B_3$ is $A_3 \setminus A_2 \cup A_1$. In general, the sequence of sets is defined by this $B_i=A_i-\bigcup_{k=1}^{i-1} A_k$ for $k=1,2,...,n$. 
Proof Attempt:
Proof (a)
Let $x\in B_i$. Since $B_i=A_i\setminus\bigcup_{k=1}^{i-1} A_k$, so by defnition of Set Difference, if something is in $B_i$, it must also be in $A_i$. Since, $x\in B_i$ implies $x\in A_i$, we conclude that $B_i\subseteq A_i$ (by definition of Subset).
Proof (b)
On which, I found an elementary approach from here.
Proof (c)
To show that $X=\bigcup_{i=1}^n B_i$, we must show that $X\subseteq\bigcup_{i=1}^n B_i$ and $\bigcup_{i=1}^n B_i\subseteq X$.
Part 1. Show that $X\subseteq\bigcup_{i=1}^n B_i$. Let $x\in X$ then $x\in\bigcup_{i=1}^n A_i$ as defined.It implies that $x$ is an element of some $A_i$'s (by definition of $\cup$).If $i_0$ is the least such value of $i$ such that $x\in A_{i_0}$ then $x\in A_{i_0}\setminus \bigcup_{k=1}^{i_0-1} A_k$ (by definition of $\setminus$). It further implies that $x\in B_{i_0}$ and hence $x\in\bigcup_{i=1}^n B_i$ by definition of $\cup$. Since $x\in X$ implies $x\in\bigcup_{i=1}^n B_i$, we conclude that $X\subseteq\bigcup_{i=1}^n B_i$ (by definition of $\subseteq$).
Part 2. Show that $\bigcup_{i=1}^n B_i\subseteq X$. Let $x\in\bigcup_{i=1}^n B_i$ then $x\in B_i$ (for some $i$). If $i_0$ is the least such value of $i$ such that $x\in B_{i_0}$ and since $B_i\subseteq A_i$ which we proved already above, it implies that $x\in A_{i_0}$ which further implies that $x\in\bigcup_{i=1}^n A_i$ (by definition of $\cup$). Hence, $x\in X$. Since, $x\in \bigcup_{i=1}^n B_i$ implies that $x\in X$, we conclude that $\bigcup_{i=1}^n B_i\subseteq X$.
Conclusion. Since, $X\subseteq\bigcup_{i=1}^n B_i$ and $\bigcup_{i=1}^n B_i\subseteq X$ then $X=\bigcup_{i=1}^n B_i$. 
Note: The proof is so elementary that fits to beginners like me.
Is this right already?

Take $B_1=A_1,B_2=A_2\setminus A_1, \cdots, B_n=A_n\setminus (A_1 \cup A_2 \cup \cdots \cup A_{n-1})$.  The idea here is if $x \in X$ then $x\in A_k$ for some $k$ and there is  smallest $k$ with this property. For that $k$ we get $x \in B_k$,
