Prove that $Z-(Y-X)=X\cup(Z-Y)$ if $X\subset Y\subset Z$ 
Let $X\subset Y\subset Z$. Prove that $Z-(Y-X)=X\cup(Z-Y)$.

Here, $A-B$ is the complement of $B$ in $A$.
To establish equality, I need to show that $Z-(Y-X)\subset X\cup(Z-Y)$ and $X\cup(Z-Y)\subset Z-(Y-X)$. Here's how I start the proof:

Let $\alpha\in Z-(Y-X)$. Then $\alpha\in Z$ and $\alpha\not\in Y-X$, which means $\color{red}{\alpha\not\in Y}$ and $\alpha\in X$. The fact that $\alpha\in X$ is sufficient to establish that $\alpha\in X\cup(Z-Y)$, and so $Z-(Y-X)\subset X\cup(Z-Y)$.

But $X\subset Y$, which would mean $\alpha\in Y$. This seems to contradict the highlighted portion above. What am I missing here?
I can see that equality holds by looking at the Venn diagram below. $X$ is green/innermost region, $Y$ is blue/middle, and $Z$ is red/outermost. Then $Y-X$ is the blue region alone, and so $Z-(Y-X)$ is made up of both the green and red regions. On the other hand, $Z-Y$ is the red region alone, so $X\cup(Z-Y)$ is that plus the green region. So it's (pictorially) true that $Z-(Y-X)=X\cup(Z-Y)$.

Based on this picture, it seems to be the case that $\alpha$ cannot belong to $X$. Then is the "contradiction" I point to above actually of no consequence to the proof?
 A: You write:

Let $\alpha\in Z-(Y-X)$. Then $\alpha\in Z$ and $\alpha\not\in Y-X$, which means $\color{red}{\alpha\not\in Y}$ and $\alpha\in X$. 

But that is wrong. If $\alpha\not\in Y-X$, then $\alpha\not\in Y$ or $\alpha\in X$
To see this, note that $\alpha \in Y-X$ iff $\alpha \in Y$ and $\alpha \not \in X$, and negating the latter, we get $\alpha\not\in Y$ or $\alpha\in X$
A: You can also use directly the De Morgan laws.
$\begin{align}
Z-(Y-X)&=Z\cap(Y-X)^\complement & \text{subtraction formula : } A-B=A\cap B^\complement\\
&=Z\cap(Y\cap X^\complement)^\complement & \text{subtraction formula} \\
&=Z\cap(Y^\complement\cup X) & \text{inversion : } (A\cap B)^\complement=(A^\complement\cup B^\complement)\\
&=(Z\cap X)\cup(Z\cap Y^\complement) & \text{distribution : }A\cap(B\cup C)=(A\cap B)\cup(A\cap C)\\
&=(Z\cap X)\cup(Z-Y) & \text {subtraction formula} \\
&=X\cup(Z-Y) & X\subset Z\implies Z\cap X=X\end{align}$
Note: we didn't use the particular condition on $Y$ being inside $Z$ and containing $X$, so the formula is a bit more general.
