Let $B$ be integral over $A$. Show that $\phi:A/Q\to B/P$ defined by $\phi(a+Q)=a+P$ is injective, where $Q=A\cap P$, and $P$ is an ideal of $B$.
What I tried was the brute force way, which is supposing $\phi(a_{1}+Q)=\phi(a_{1}+Q)$. Then $a_{1}+P=a_{2}+P$. Then $a_{1}=a_{2}+b$ for some $b\in P$.
Is this going the right direction though, or is there anything simpler? I am not sure what to do with the next step...
Brute force is fine. Suppose $\phi(a+Q)=\phi(b+Q)$. Then $a+P=b+P$, so $a-b\in P$. Hence $a-b \in P\cap A$, since $a,b\in A$. Therefore $a-b\in Q$, so $a+Q=b+Q$.
You're essentially proving that if $\phi:A\to B$ is a ring homomorphism, then the induced map $A/\ker\phi \to B$ is injective. Here $B=B/P$, and $\phi:A\to B/P$ is the composite $A\hookrightarrow B \to B/P$. You can check that here $\ker\phi = Q$.