Discrete Math (Proving by contradiction for sets) So, the question is:
Prove by contradiction that if A$\cap$B$\subseteq$C and x$\in$A-C, then x$\notin$B.

So My process is
Let

A:{X,Y,Z}

B:{Z,Y,V}

C:{Y,Z,Q}

A$\cap$B: {Z,Y}

A-C = {X,Y,Z} - {Y,Z,Q} = {X}

So it follows that x$\in$A-C but x$\notin$B.

So I'm not sure if this is the proper way to prove by contradiction as I have never done contradiction proofs with sets before
To prove $P \to Q$ by contradiction, you must assume both $P$ and $\lnot \, Q$ (not $Q$) are true and show that that leads to a contradiction.

In your particular case, you should start by assuming that $(A\cap B)\subset C$ and $x \in (A-C),$ as well as $x \in B.$ Then the last two conditions imply that $x \in (A \cap B).$ Can you see how to derive the contradiction from there?
Thanks everyone I got the answer

We assume that:

x∈A-C and x∈B

If, x∈A-C then x∈A

If, x∈A-C then x∉C

If, x∈B then X∈A∩B

But if X∈A∩B, then X∈C, since A∩B⊆C which is not possible, since x∉C

Therefore by contradiction x∉B.