Some Background and Motivation: In this question, it is shown that an integral domain $D$ such that $F \subset D \subset E$, $E$ and $F$ fields with $[E:F]$ finite, is itself a field. However, a significantly more general result holds and seems worthy, of independent address; hence,
Let
$F \subset E \tag 1$
be fields with
$[E:F] < \infty; \tag 2$
if $R$ is a ring such that
$F \subset R \subset E, \tag 3$
show that $R$ is in fact a field.
If $x\in R\setminus F$, then for some minimal $n\in\mathbb{N}$, $x$ is a root of a polynomial $\sum_{i=0}^nc_ix^i$ over $F$ of degree $n$. Otherwise, $\{1,x,x^2,\ldots\}$ is a basis of an infinite dimensional vector space over $F$, but $[E:F]$ is finite. And note that $c_0\neq0$. Otherwise $x$ would be a zero-divisor, and this all happens within field $E$.
So for any $x\in R\setminus F$, you have $x\cdot\overbrace{\left(-\frac{1}{c_0}\sum_{i=1}^{n}c_ix^{n-1}\right)}^{\in R}=1$. So $R$ contains the inverse of $x$.