The following is a proposition from abstract algebra. The book gives this as a proposition and asks the reader to do it as an exercise. It seems it should be easy. (Assuming every ring has a $1$.)
Let $R$ be a subring of $S$. If $S \setminus R$ is closed under multiplication, then $R$ is integrally closed in $S$.
I went by contradiction. Let $s$ be in $R \setminus S$. Then, we get that $$ s^n+a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}=0 $$ for some $n$ and the $a$'s are in $R$. Now I am thinking this $$ s^n=(-1)(a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}) $$ Now multiply both sides by $s$ again to get $$ ss^n=(-s)(a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}) $$ Now by closure, $ss^n$ is in $S \setminus R$ which means the rest $a_1s^{n-1} + \cdots + sa_{n-1}+a_{n}$ is in $S \setminus R$ which means the $a_i$'s are in it, which is a contradiction. Does this look correct?
Edit (not by OP): The above is also Exercise 7 from Chapter 5 of Atiyah and Macdonald's Introduction to Commutative Algebra.
Hint $\ $ Choose $\,n\,$ minimal. Then $\ s f(s) = s (s^{n-1}+ r_{n-1} s^{n-2} + \cdots + r_1)\, =\, -r_0.\,$ But $\,f(s)\not\in R\,$ by minimality of $\,n\,$ so $\,s,f(s)\in S\backslash R \,\Rightarrow\, sf(s) = -r_0\in S\backslash R ,\,$ contradiction.