$A \subseteq B \subseteq C ; A' \subseteq C' ; |A|=|A'| , |C|=|C'|$ ; then $\exists B' $ s.t. $A' \subseteq B' \subseteq C' $ , $|B|=|B'|$? Let $X$ be a non-empty set and $A,B,C,A',C' \in \mathcal P(X)$ be such that $A \subseteq B \subseteq C ; A' \subseteq C'$ and 
$|A|=|A'| , |C|=|C'|$ ; then is it true that $\exists B' \in \mathcal P(X) $ such that $A' \subseteq B' \subseteq C' $ where $|B|=|B'|$ 
? ( here by $|A|=|B|$ we mean there is a bijection between $A$ and $B$ ) 
 A: Of course you can't simply conclude that $C\setminus A$ has the same cardinality as $C'\setminus A'$. Take as a counterexample $C=C'=\mathbb{N}$, $A=\mathbb{N}\setminus\{0,1\}$ and $A'=\mathbb{N}\setminus\{0\}$.
The answer is not quite trivial and needs some cardinal arithmetic. So let's consider the case where $A$ and $C$ are finite first. I think then the answer is quite obvious.
Now let's assume that $\vert A \vert<\vert C \vert$, i.e. the cardinality of $A$ is strictly smaller than the cardinality of $C$ and $\vert C \vert$ is infinite. There is a law for cardinal numbers $\alpha,\beta$ where at least one of them is infinite which says $\alpha+\beta=\max\{\alpha,\beta\}$. You can find a proof here.  
From this it follows that $\vert C\vert=\vert C\setminus A \vert+\vert A \vert =\max\{\vert C\setminus A\vert,\vert A\vert\}$. And since $A$ was assumed to have strictly smaller cardinality you can conclude that $\vert C\setminus A\vert=\vert C \vert$. And the same holds on the other side and therefore you can make the following argument:
This means there is a bijection $\varphi: C\setminus A\to C'\setminus A'$ and also a bijection $\phi: A \to A'$. 
Now construct $f:C \to C'$ by
$$ f(c)=\begin{cases}\varphi(c) \qquad \forall c \in C\setminus A\\
\phi(c) \qquad \forall c \in A  \end{cases} $$
This is a bijection and $B':=f(B)$ is the set you are looking for since $\vert B\vert=\vert B' \vert$ via $f$ and also $A'\subset B' \subset C'$ by construction.

Now consider the case where $\vert A\vert =\vert C\vert$, then any set $B$ with $A\subset B\subset C$ has to have the same cardinality and you can take $B'=A'$. If you want strict inclusions you can come up with counterexamples. Take:
$$ A=\mathbb{N}\setminus\{0,1\}, B=\mathbb{N}\setminus\{0\}, C=\mathbb{N} $$
and
$$A'=\mathbb{N}\setminus\{0\}, C'=\mathbb{N}. $$
Then $A$ and $A'$ have the same cardinality and so have $C$ and $C'$ but you cant find a $B'$ strictly between $A'$ and $C'$.
