Let $E/F$ be a field extension and $F\subset R\subset E$ where $R$ is a subring of $E$. I want to show that: $E/F$ is algebraic iff every subring $R$ of $E$ which contains $F$ is a field.
I already proved that if $E/F$ is algebraic then $R$ is a field. (Note that $R$ is completely arbitrary).
Now I am stuck at the converse, namely, if for every subring $R$ it holds that it is a field ($R$ given as above), then $E/F$ is algebraic.
My strategy would be that I choose an arbitrary $r\in R$ and look at $F[r]=F(r)$ where I have used that every subring is a field. But now I have trouble proceeding, because I can not work with finite polynomials to show anything.
If $E$ is not an algebraic extension of $F$, then there's some $\omega\in E$ which is transcendental over $F$. Take $R=F[\omega]$. It is a ring, but not a field ($\omega$ has no inverse there).