If $A \subset Y \subset X$ with $A'\subset X$ and $A = A' \cap Y$, then $A = A'$ 
If $A \subset Y \subset X$ with $A'\subset X$ and $A = A' \cap Y$, then $A = A'$

Is the statement in the title correct?  I don't think so, but my attempt seems convincing it is:
$$\begin{align}A = A' \cap Y &\implies Y-A ~{= (Y-A') \cup (Y -Y)  \\= Y -A'} \\ & \implies Y-(Y-A) = Y-(Y-A') \\ & \implies A = A'\end{align}$$
Where is my mistake?
 A: It is true that$$ Y \setminus A = Y \setminus A^\prime ,$$ so that  $$ Y \setminus (Y \setminus A) = Y\setminus (Y \setminus A^\prime) . $$
However, expanding the left-hand side:
$$ Y \setminus (Y \setminus A) = Y \cap (Y \cap A^c)^c = Y \cap (Y^c \cup A)= Y \cap A.$$
Similarly, the right-hand side is $Y \cap A^\prime$.
In other words, you've proven that $$ Y \cap A = Y \cap A^\prime, $$ which is true. However, this does not imply that $A=A^\prime$.
A: Well $Y\smallsetminus (Y\smallsetminus A) = Y\smallsetminus (Y\smallsetminus A^\complement)$ does not imply that $A=A^\complement$, only that $Y\cap A^\complement = Y\cap A$, which you already knew, redering the 'proof' moot. 
Also...

Assume $A=A^\complement\cap Y$ .   That means every element of $A$ is an element of $A^\complement\cap Y$.   However, no element of $A$ can be an element $A^\complement\cap Y$.   Thus $A$ is empty.   This means $A^\complement\cap Y$ is empty too, which mandates that $Y$ is empty also (since $Y=(A^\complement\cap Y)\cup(A\cap Y)$.)
$$A=A'\cap Y ~\to~( A=\emptyset~\wedge~ Y=\emptyset)$$
