Suppose $F$ is a subfield of a field $K$, and let $R$ be a ring contained in $K$ and containing $F$. Can we conclude that $R$ is a subfield of field $K$?
The answer is "yes" in case the extension is finite or more generally algebraic. See: Ring Inside an Algebraic Field Extension
Is this still true without that condition?
No:
Consider the case in which $F = \Bbb Q$, the rationals, and $K = \Bbb R$, the reals; let $\tau$ be any transcendental real number, e.g. we might take $\tau = e$ or $\tau = \pi$. Let $R$ be the ring $\Bbb Q [\tau]$, i.e. polynomial expressions in $\tau$ with rational coefficients. Then $\Bbb Q \subset R \subset \Bbb R$. $R$ is easily seen to be a subring $\Bbb R$, but it cannot be a field; it does not contain $\tau^{-1}$; if it did, we would have, for some polynomial $p(x) \in \Bbb Q [x]$,
$p(\tau) = \tau^{-1}$;
but then
$\tau p(\tau) = 1$,
so $\tau$ would satisfy the polynomial equation
$\tau p(\tau) - 1 = 0$,
which would imply that $\tau$ is algebraic over $\Bbb Q$, a contradiction. QED.