If $A \perp C$, how would one go about proving that:
$P(A,C|B) = P(A|B)P(C|B)$
I have tried:
$\frac{P(A,B,C)}{P(B)}=\frac{P(A,B)}{P(B)}\frac{P(C,B)}{P(B)}$ but I am unsure of a rule that would let me combine $P(A,B)$ and $P(C,B)$. I know that since $A \perp C$, then $P(A,C) = P(A)P(C)$ but I can't find a place to apply it.
One would not prove it. It is not necessarily true that when events $A,C$ are independent, then they will also be conditionally independent given a third event.
Witness for the Contrary: Toss two independent fair coins, one bronze and the other silver. Let event $A$ be that the bronze coin shows head, event $C$ that the silver coin does so, and event $B$ that both coins show the same face. We see that $\mathsf P(A,C)=\mathsf P(A)~\mathsf P(C)$, but on examining the conditionals:
$$\mathsf P(A\mid B)=\tfrac 12, \mathsf P(C\mid B)=\tfrac 12, \mathsf P(A,C\mid B)=\tfrac 12$$