Proving equality between sets (elementary set theory) 
Prove that: $$A \cup B = A \cup C \text{ and } A \cap B = A \cap C \implies B = C$$

Proof: Suppose  $A \cup B = A \cup C \text{ and } A \cap B = A \cap C$ but $B \neq C$. Then either $B \not\subset C$ or $C \not\subset B$. 
Let $w \in B$ such that $w \notin C$. Then $w \in A \cup B$, which implies, by hypothesis, that $w \in A \cup C$. But $w \notin C$, so it must be that $w \in A$, therefore we have that $w \in A\cap B$, which again implies that $w \in A \cap C$, a contradiction. Therefore no such $w$ exists and $B \subset C$.
Second case:
Let $w \in C$ such that $w \notin B$. Then $w \in A \cup C$, so $w \in A \cup B$
. Since $w \notin B$, then, similarly to the first case, we have $w \in A \cap C$, which again implies $w \in A \cap B$, a contradiction. Therefore $C \subset B$.
Since $B \subset C$ and $C \subset B$, then $B = C$. 
Unfortunately I'm just getting the hang of this now, so I still must defer to other people to check my work. Is everything correct? Could my writing or anything else be improved? Any help would be appreciated.
 A: In each case, you cant say $B = C$. What youve shown is that in case 1 if $w \in B$ then $w \in C$, so you've shown that $B \subset C$. Vice versa for the second case.
This also seems kind of weird to me "Then either $B \not\subset C$ or $C \not\subset B$. Both cases lead to a contradiction, as I prove here"
I would just leave that out and then show like you did that $B \subset C$ and $C \subset B$ which means $B = C$
A: Other than what XRBtoTheMoon already pointed out, your proof is fine; it's well organized, and your level of detail is just about right. In fact, for someone who is starting to do thse kinds of proofs, you are doing very well!
Of course, if you're allowed to use algebra, all English can be completely avoided:
$$B = B \cap (A \cup B) = B \cap (A \cup C) = (B \cap A) \cup (B \cap C) =$$
$$ (A \cap B) \cup (B \cap C) = (A \cap C) \cup (B \cap C) = $$
$$(A \cup B) \cap C = (A \cup C) \cap C = C$$
A: Here is a solution using indicator functions. The proof uses the fact that the map
$
A\to I_A$ is a bjection from $P(\Omega)$ to $2^{\Omega}$. Note that
$$
A\cup B=A\cup C\iff I_{A\cup B}=I_{A\cup C}\iff I_{A}+I_{B}-I_{A}I_{B}=I_{A}+I_{C}-I_{A}I_{C}\tag{1}
$$
Next the assumption
$$
A\cap B=A\cap C\iff I_{A\cap B}=I_{A\cap C}\iff I_{A}I_{B}=I_{A}I_{C}\tag{2}
$$
Thus with equation (2), equation (1) implies that
$$
I_{B}=I_{C}\implies B=C.
$$
