If $R$ is a commutative ring with a subring $S$ of finite index, it is not hard to show that there is an ideal $I$ of $R$ with $[R:I]$ finite and $I\subset S$. The idea is to take $I = \{s\in S : rs\in S, \ \forall r\in R \}$, and then observe that $I$ is the kernel of the natural map $S\to End(R/S)$ taking $s\mapsto ([r]\mapsto [rs])$, where $R/S$ is just an abelian group and $End(R/S)$ is its endomorphism ring. Note that $S$ doesn't have to contain a multiplicative unit; we just need that $S$ is closed under addition and multiplication.
I wanted to know if the same result actually holds true for non-commutative rings: if $R$ is a (unital if necessary) non-commutative ring with a subring $S$ of finite index, is there a 2-sided ideal $I$ in $S$ whose index in $R$ is finite?
The idea I had had was to define $I_R = \{s\in S: rs\in S, \ \forall r\in R \}$ and similarly $I_L = \{s\in S: sr\in S, \ \forall r\in R \}$; these are 1-sided ideals in $R$ contained in $S$, of finite index in $R$, by a similar argument to the above. We could then say that $I_R\cap I_L$ is finite index in $R$, but the intersection of left ideal and right ideal is not in general a 2-sided (or even 1-sided) ideal. Perhaps instead I should use the set $J=\{ s\in S: rsr'\in S \ \forall r, r'\in R\} $, which is a 2-sided ideal contained in $S$, but I'm uncertain it needs to have finite index.
Turns out the answer is yes. This is lemma 1 in Lewin, Jacques. Subrings of finite index in finitely generated rings. J. Algebra 5 (1967), 84--88.