Use direct proof to prove: If $A \cap B = A \cap C$ and $A \cup B = A \cup C$, then $B = C$ I'm interested in knowing if the method I used is correct. I've been teaching myself proofs lately and I am having difficulties with how to approach a problem so any general tips would be awesome as well! Here is what I have:

Assume $A \cap B = A \cap C$ and $A \cup B = A \cup C$.
Let $x \in B$.
Let $x \in A$.
Then, $x \in A \cap B$.
Since $A \cap B = A \cap C$, $x$ must also be an element of $C$.
That is, $x \in C$.

Using the same thinking we can prove the $A \cup B = A \cup C$ case. Ie.

Assume $A \cap B = A \cap C$ and $A \cup B = A \cup C$.
Let $x \in B$.
Let $x \in A$.
Then, $x \in A \cap B$.
Since $A \cup B = A \cup C$, $x$ must also be an element of $C$.
That is, $x \in C$.

Finally, we see that for any $x \in B$, it follows that $x \in C$ as well. We can show that the converse is true by letting $x \in C$. Thus, $B = C$.
 A: The first is almost okay, but the second subproof is erranous.

Your first proof should demonstrate that: "Since $A\cap B=A\cap C$ any element of $B$ that is an element of $A$ is also an element of $C$.   Hence we infer $A\cap B\subseteq C$ and symmetrically $A\cap C\subseteq B$"
Your second proof needs to demonstrate that: "Since $A\cup B=A\cup C$ then any element of $B$ that is not an element of $A$ is an element of $C$.   Therefore $B\setminus A\subseteq C$ and symmetrically $C\setminus A\subseteq B$"
Then you need to add something like: "Thus all of the elements of $B$ are elements of $C$; that is: $B\subseteq C$, and likewise all of the elements of $C$ are elements of $B$; that is: $C\subseteq B$.   Therefore these two sets are identical."
$$A\cap B=A\cap C, A\cup B=A\cup C ~\implies~ B=C$$
$\Box$
In your own words, of course.
A: You haven't proven that if $x\in B$ then $x\in C$. Rather, you've proven that if $x\in B$ and $x\in A$, then $x\in C$. 
You have fallen into a common error in your usage of the word "let." I've seen this usage a lot, but I'm not sure what you (or others) mean by it. I think most mathematicians would use the word "assume $x\in B$ and $x\in A$." 
Here's one approach: Prove that for any $A,B$:
$$B=\left((A\cup B)\setminus A\right)\cup (A\cap B))$$ 
Then if $A\cup B=A\cup C$ and $A\cap B=A\cap C$ then:
$$\begin{align}B&=\left((A\cup B)\setminus A\right)\cup (A\cap B))\\
&=\left((A\cup C)\setminus A\right)\cup (A\cap C))=C
\end{align}$$
A: I believe the proof should go like this:
Assume $x\in B$. Then $x\in A \cup B=A\cup C$. If $x\in C$ then good. If $x\in A$ then it is in $A\cap B=A\cap C$ so it is in $C$.
Then assume $x\in C$. By a symmetric argument to the one just given, it is in $B$.
