$A \cap B \subset (A \cap C) \cup (B \cap C')$ How do i show this? 
$A \cap B \subset (A \cap C) \cup (B \cap C')$ 
$$ x \in A \cap B$$
$$ \implies x \in A \cap B \cup \emptyset$$
$$ \implies x \in A \cap B \cup (C \cap C')$$
$$ \implies x \in (A \cap B \cup C)\cap(A \cap B  \cup C')$$
How do i proceed?
 A: NOTE: This answers the previous version of this question.
I would use a different approach.
Lemma: $P\subseteq P\cup Q$
Proof:
$$\begin{align}
x\in P & \implies x\in P \lor x\in Q \\
 & \implies x\in P\cup Q
\end{align}$$
Proof of main problem:
In the lemma, substitute $P=A\cap B$, $Q=B\cap C'$. Then we get
$$A\cap B\subseteq (A\cap B) \cup (B\cap C')$$

If you really want to continue the approach you started (which actually uses my lemma without being explicit),
$$\begin{align}
x\in A\cap B
 &\implies x\in A\cap B \lor x\in B\cap C' \\
 &\implies x\in (A\cap B) \cup (B\cap C')
\end{align}$$
A: Use the fact that 
$$X \subseteq (X \cap C) \cup (X \cap C')$$
Then
$$A \cap B \subseteq (A \cap B \cap C) \cup (A \cap B \cap C')\subseteq (A  \cap C) \cup ( B \cap C')$$
A: Suppose that $x$ is in both $A$ and $B$, i.e. $x \in A \cap B$. Then either $x$ is in $C$, in which case it is then in the intersection $A \cap C$; or it's not in $C$, in which case it is in $C'$, thus it is in $B \cap C'$. In the end, $x$ is in both cases in the union $(A \cap C) \cup (B \cap C')$. Since this was true for all $x \in A \cap B$, it follows that $A \cap B \subset (A \cap C) \cup (B \cap C')$.
(This answer is here because I don't believe math should be about wildly manipulating strings of symbols using arcane rules in hopes of getting the correct sequence of symbols at the end of the implication arrow.)
