$A \subseteq B$ if and only if $B' \subseteq A'$? I have already tried to prove it in my own way and would like to share my results in hope that a mathematician somewhere can tell if I am correct. I believe that the proof consists of 2 parts: First, prove left to right that $A \subseteq B \implies B' \subseteq A'$. Second, prove right to left that $B' \subseteq A' \implies A \subseteq B$. 
In the first part, we assume $A \subseteq B$. We have $x \in A' \implies x \not \in A$ and since $A \subseteq B$, then $x \not \in B$. On the other hand, $x \in B' \implies x \not \in B$ and $x \in A'$ therefore $A' \cap B' = B'$ which means $B' \subseteq A'$. 
The second part of the proof is more or less the same. Is my approach correct? Thanks.
 A: Your argument is rather unclear and not quite correct. What you want to do is first assume $A \subset B$ and let $x \in B'$. Then, since $x \not\in B$ and since $A \subset B$, we have that $x \in A'$. So $B' \subset A'$. The other direction follows from the previous implication since $A'' = A$ and $B'' = B$ so $B' \subset A' \implies A \subset B$
A: Suppose $A\subset B$. Then $x\in A \implies x\in B$. Suppose that $x\in B^C$, but $x\not\in A^C$, then $x\in A, x\not\in B$, this implies $A\not\subset B$, a contradiction. So $ B^C \subset A^C.$
Suppose conversely that $B^C\subset A^C$, and that $\exists x: x\in A, x\not\in B$. Then $x\in B^C, x\not \in A^C$ and so, $B^C\not\subset A^C$ a contradiction. So $A\subset B$.
This shows both directions of the equivalence and completes the proof.
A: Let "$p$" be the  proposition "$x\in A$", and let "$q$" be the  proposition "$x\in B$".
 $(A\subseteq B)$ is equal to "$\forall x, (x\in A \implies x\in B)$". Similarly $(B'\subseteq A')$ is equal to "$\forall x, (x\notin B \implies x\notin A)$". Now use Truth Table below:
\begin{array}{c|c|c|c|c}
p & q  & \lnot p &\lnot q & p \Rightarrow q  & \lnot q \Rightarrow \lnot p\\
\hline
T & T & F &F & T & T\\
\hline
T & F & F & T & F & F\\
\hline
F & T &T &  F& T &T\\
\hline
F & F &T  & T& T &T\\
\end{array}
