If $R$ is integral over $S$, then $\operatorname{Frac}(R)/\operatorname{Frac}(S)$ is finite extension of fields How to show that: 

If $R\supset S$ are integral domains, $R$ is integral over $S$, $K$ and $L$ the fraction fields of $R$ and $S$ respectively, then $K/L$ is a  finite extension of fields.

 A: You can't hope to get a finite field extension as long as there are algebraic field
extensions which are not finite, e.g. $\mathbb Q\subset\mathbb Q(\sqrt 2,\sqrt[4] 2,\sqrt[8] 2,\dots)$. Instead one can prove that $L\subset K$ is algebraic.
Let $x\in K$, where $K$ is the field of fractions of $R$. We want to prove that $x$ is algebraic over $L$ (the field of fractions of $S$). But $x=a/b$ with $a,b\in R$. Since $a,b$ are integral over $S$ they are algebraic over $L$, so their quotient is algebraic over $L$.
A: First, you need to assume that $R$ and $S$ are integral domains. Second, I will prove a more general claim:

Let $A$ and $B$ be rings and $S \subset A$ multilicatively closed subset of $A$. Assume $B$ is integral over $A$. Then $S^{-1}B$ is integral over $S^{-1}A$.

Proof. Let $x/s \in S^{-1}B$ with $x \in B$ and $s \in S$. Let 
$$ x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = 0 $$
be the polynomial equation that $x$ satisfies (since $B$ is integral over $A$). Then $x/s$ satisfies
$$ x^n/s^n + a_{n-1} x^{n-1}/s^n + ... + a_1 x/s^n + a_0/s^n = 0 $$
which is
$$ (x/s)^n + (a_{n-1}/s) (x/s)^{n-1} + ... + (a_1/s^{n-1}) (x/s) + (a_0/s^n) = 0 $$
with $a_{n-1}/s , ... , a_0/s^n \in S^{-1}A$ so $x/s$ is integral over $S^{-1}A$. 
$\quad \square$
