Let $A,B,C$ are the subsets of $E$. Prove that: If $A\cup C\subset A\cup B$ and $A\cap C\subset A\cap B$ then $C\subset B$. Let $A,B,C$ are the subsets of $E$. Prove that: If $A\cup C\subset A\cup B$ and $A\cap C\subset A\cap B$ then $C\subset B$.

My attempt:
For $A\cup C\subset A\cup B$ then with all $x\in C\subset A\cup B$ $(1)$.
For $A\cap C\subset A\cap B$, we have $x\in C\Rightarrow x\in B$ $(2)$.
From $(1),(2)$ we have $C\subset B$.
I think my solution is not good, please correct me if I am wrong. Thank you so much!
 A: As pointed out in the comments, your attempt is wrong. Essentially, you claim that step $(2)$ is true without giving a proof, and actually $(2)$ doesn't hold: $A∩C⊆A∩B$ does not imply that if $x∈C$ then $x∈B$.
As a counterexample, take $A = \{0\}$, $B = \{0,1\}$ and $C = \{2\}$.
A correct proof is the following. Suppose that $A \cup C \subseteq A \cup B$ and $A \cap C \subseteq A \cap B$. We want to prove that $C \subseteq B$.
Let $x \in C$. There are two possibilities:

*

*either $x \in A$, then $x \in A \cap C$ and so $x \in A \cap B$ (because $A \cap C \subseteq A \cap B$), therefore $x \in B$;

*or $x \notin A$, then $x \in A \cup C$ (since $x \in C$) and so $x \in A \cup B$ (since $A \cup C \subseteq A \cup B$), and more precisely $x \in B$ because $x \notin A$.

Thus, we proved that in any case $x \in B$, under the assumption that $x \in C$.
Hence, $C \subseteq B$.
A: $A\cap C\subseteq A\cap B$ does not in itself imply $x\in C\implies x\in B$ i.e. $C\subseteq B$. Take, for example, $A=\{1\},C=\{2\},B=\{\}$. So your conclusion $(2)$ is wrong.
Note that $A\cap C\subseteq A\cap B\subseteq B$ does imply that $A\cap C\subseteq B~(2)$.
Now, if $x\in C$ there could be two possibilities:

*

*$x\in C\cap A$: By $(2)$ we get $x\in B$.

*$x\in C\cap A^C$: By $(1),x\in A\cup B$. But $x\notin A$, so we get $x\in B$.

This gives $x\in C\implies x\in B$ i.e. $C\subseteq B$.

Alternatively, $(1)$ gives $C\cap A^C\subseteq(A\cup B)\cap A^C=B\cap A^C$.
$(2)$ gives $C\cap A\subseteq B$.
Taking unions,$$C=(C\cap A)\cup(C\cap A^C)\subseteq B\cup(B\cap A^C)=B$$
