How to prove that if $A\cap B' =\varnothing$ then $A \subseteq B$? What are the steps of proving this?

If $A\cap B' = \varnothing$ then $A \subseteq B$

where $B'$ is the complement of $B$.
 A: $x\in A \implies x\notin B^c \implies x\in B$
A: Let me give it a try: 


*

*For every $A$, $A=b(A)\cup c(A)$ with $b(A)=A\cap B\subseteq B$ and $c(A)=A\cap B'\subseteq B'$. 
To prove this, note that $B\cup B'$ is everything hence $A=A\cap(B\cup B')=(A\cap B)\cup(A\cap B')$ by distributivity of $\cup$ with respect to $\cap$.

*The hypothesis is that $c(A)=\varnothing$. Hence $A=b(A)$. Recall that $b(A)\subseteq B$, always. Hence $A\subseteq B$.


Note: The conditions that $A\cap B'=\varnothing$ and that $A\subseteq B$ are in fact equivalent.
A: Nana proves it directly. A proof by contradiction is supplied here.
If $A = \emptyset$, then $A \subseteq B$.
If $A \neq \emptyset$, let $x \in A$. We want to show that $x \in B$.
Suppose not, then $x \in B'$. We have $x \in A \cap B' = \emptyset$ which is not possible.
Hence $A \subseteq B$.
Edit: I answered this question to test the proof approach and technique I learnt from school years ago. Would anybody point out anything that is wrong or inappropriate in the proof so that I can improve them? Thanks.
