Proving Set Equality $A=B$

Question

I have the following question:

$$\text{Suppose $A,B,C$ are sets so that $A \cap C = B \cap C $ and $A \cup C=B \cup C$. Prove that $A=B$.}$$

My question is that the way I am doing it seems rather straightforward to me, but my professor has a much different approach.

Here is my approach:

$$\text{Let $x \in A$}. \text{Then $x \in A\cap C$},\text{so $x \in B \cap C$}. \text{Hence $x \in B$ } \text{and $A \subseteq B$}.\\ \text{Let $x \in B$. Then $x \in B \cap C $, so $x \in A \cap C$. Hence $x \in A$ and $B \subseteq A$.}\\ \text{Therefore, $A=B$.}$$

Here is my professor's approach:

$$A=(A \cap C) \cup (A \cap C^c)\\B=(B \cap C) \cup (B \cap C^c)\\\text{Now}\; (A\cup C)\cap C^c = A\cap C^c \text{ and } (B\cup C) \cap C^c = B \cap C^c \\\text{so } A=B$$

I was wondering if there is any issue in my approach that would explain why he answered the problem in this way instead? Thank you!

Answer 1

The implication $x\in A \to x \in A\cap C$ is false. Why $x\in C$?

Answer 2

Would this work instead,

$$x \in A \rightarrow x \in A\cup C \rightarrow x\in B\cup C\\\text{so we either have that:}\\(1)\;x\in B , x\notin C\\(2)\;x\in C, x\notin B \\(3)\;x\in B, x\in C\\ \text{For (1) we have that $x \in B$. For (3) we have that $x \in B$. So we would just need to worry about (2).} \\\text{(2) States $x \in C$. Therefore, $x\in C$ and $x\in A$. So $x\in A\cap C.$ But then $x \in B\cap C$, so $x \in B$. But $x\notin B$.}\\\text{So only (1) and (3) could hold. In which case, $x \in B$ and $A \subseteq B.$}$$

And it would be the same type of logic for $B \subseteq A.$ Would this work instead?