The contraction of $B_p\cdot q$ is $p\cdot A_p$ I am trying to prove the following:

Let $A\subset B$ with $B$ integral over $A$. Let $q\subset q' \subset B$ be prime ideals. If $q^c=q'^c\subset A$ then $q=q'$.

Here is my attempt with the points I am confused about inside brackets:
Let $p=q^c$ and take the localisation at $p$. Then $p\cdot A_p$ is a maximal ideal in $A_p$.
Then it should be that $p\cdot A_p$ is the contraction of $q \cdot B_p$ (I'm not sure how to see this).
Since $p\cdot A_p$ is maximal it gives that $q \cdot B_p$ will be maximal.
This gives that $q'\cdot B_p$ is either $q \cdot B_p$  or all of $B_p$. (I can't see why this cannot be all of $B_p$).
Now if we know that $q\cdot B_p=q\cdot B_q'$ (how can we lift this to get that $q=q'$).
 A: I change your notation slightly for convenience; in particular, I let $P$, $Q$, and $Q'$ denote what you refer to as $p$, $q$, and $q'$. Also, let $S=A\setminus P$.
For your first question, we wish to show that $P(S^{-1}A)=Q(S^{-1}B)\cap(S^{-1}A)$. The inclusion $\subseteq$ is clear, and for the other direction consider an arbitrary element $\tilde{q}$ in the right hand side. There must be some $q\in Q$, $a\in A$, $s_1,s_2\in S$ such that $\tilde{q}=q\big/s_1=a\big/s_2$. Thus, there is $t\in S$ such that $(as_1-qs_2)t=0$, so we have $as_1t=qs_2t\in Q$. Clearly also $as_1t\in A$, so in fact $as_1t\in Q\cap A=P$. Since $P$ is prime and $s_1,t\notin P$, this means that $a\in P$, whence $\tilde{q}=a\big/s_2\in P(S^{-1}A)$.
For your second question, suppose that $Q(S^{-1}B)=S^{-1}B$. Then there are $q\in Q$ and $s\in S$ such that $q\big/s=1\big/1$, and hence there is $t\in S$ such that $(s-q)t=0$. But this means $st=qt\in Q$, and again clearly $st\in A$, so – as above – this means $st\in Q\cap A=P$, a contradiction.
For your final question, let $q'\in Q'$ be arbitrary. Now, we have $$q'\big/1\in Q'(S^{-1}B)=Q(S^{-1}B),$$ and so there are $q\in Q$, $s\in S$ such that $q'\big/1=q\big/s$. This means there is some $t\in S$ such that $(q's-q)t=0$, whence $q'st=qt\in Q$. Since $st\in A\setminus P$, we have $st\notin Q$, and so since $Q$ is prime this means $q'\in Q$, as desired.
