Wanted to see if my proof of this theorem was correct. If it is not, please correct me!
Theorem. Suppose $ A \subseteq C$, and $B$ and $C$ are disjoint. Prove that if $ x \in A $ then $ x \notin B $.
Proof. Suppose $ x \in A $, $ A \subseteq C $, and $B$ and $C$ are disjoint. Then $A$ and $B$ are disjoint, since $A$ is a subset of $C$. Suppose $x \in B$. This contradicts the fact that $x \in A$, because $A$ and $B$ are disjoint. Thus, if $x \in A$ then $x \notin B$.
I think you have all the right intuitions. You might consider wording your proof differently just to improve clarity, but this is absolutely minor and the idea behind your proof is essentially correct. Also, you don't need to deduce that $A \cap B = \emptyset$ really. You can still derive a contradiction without that. Here is an example:
Proof: Suppose $x \in A$. Since $A \subseteq C$ one has $x \in C$. Suppose towards contradiction that $x \in B$ as well. Then $x \in B \wedge x \in C$ or $x \in B \cap C$. This contradicts our assumption that $B \cap C = \emptyset$. Thus $x \notin B$.