WTS: $X = (X \cap (C \setminus A)) \cup (X \cap B)$ Let $A,B \subset C$, with $A \subset B$. I want to show that for every subset $X$ of $C$, it follows: $X = (X \cap (C \setminus A)) \cup (X \cap B)$. Let
\begin{equation}\begin{split}
&~x \in (X \cap (C \setminus A)) \cup (X \cap B) \\
\Leftrightarrow&~x \in X \wedge (x \in C \setminus A \vee x\ \in B) \\
\Leftrightarrow&~x \in X \wedge~...\color{red}?... \\
\Leftrightarrow&~x \in X \wedge x \in C \\
\Leftrightarrow&~x \in X.
\end{split}\end{equation}
 A: You approach is fine, let me try and help you fill the gap in your proof.$
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\newcommand{\endcalc}{\end{align}}
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$
You started out perfectly fine: calculate which $\;x\;$ are elements of the set on the right hand side, and then work towards $\;x \in X\;$.
$$\calc
    x \in (X \cap (C \setminus A)) \;\cup\; (X \cap B)
\op=\hint{definitions of $\;\cup, \cap, \setminus\;$}
    (x \in X \land x \in C \land x \not\in A) \;\lor\; (x \in X \land x \in B)
\op=\hint{$\;x \in C\;$ follows from $\;x \in X\;$ by the assumption $\;X \subseteq C\;$}
    (x \in X \land x \not\in A) \;\lor\; (x \in X \land x \in B)
\op=\hint{logic: extract common conjunct}
    x \in X \;\land\; (x \not\in A \lor x \in B)
\op=\hint{logic: rewrite -- to make the next step easier to see}
    x \in X \;\land\; (x \in A \then x \in B)
\op=\hint{RHS is true by the assumption $\;A \subseteq B\;$; simplify}
    x \in X
\endcalc$$
Therefore, by set extensionality, $\;(X \cap (C \setminus A)) \;\cup\; (X \cap B) \;=\; X\;$.
A: Hint (1):$\cap$ and $\cup$ are distributive over each other in the style of the distributive law of arithmetic. That is :  $p\cap (q\cup r)=(p\cap q)\cup (p\cap r)$ and $p\cup (q\cap r)=(p\cup q)\cap (p\cup r)$. Hint (2): $p\cap (q\backslash r)=(p\cap q)\backslash (p\cap r)$.
