Commutative ring $R\neq \{0\}$ in which every subring is finite is a field question on proof 
Assume we are given a commutative ring $R \neq \{0\}$ with no zero
  divisors (not necessarily with a unit element) in which every proper subring only has finitely many elements. Show
  that $R$ is a field.

I am aware of the fact that there is a similar question found here, but my questions are of a different nature. First of all, the formulation "...in which every subring only has finitely many elements..." could be simply substituted with $R$ is finite. Am I right?
Then in the prove, we construct a mapping $$\rho_y : \begin{cases} R \to (y)\\x \mapsto xy\end{cases}$$ for some $y \in R\setminus \{0\}$. This is clearly a bijection (kernel is trivial by cancellation law). By $(y) \subseteq R$ we get that $R = (y)$. In the solutions in my book it is argumented that this is since $(y)$ is finite by assumption. But does this not also hold if $R$ or any of its subrings is necessarily finite?

Assuming you're talking about possibly nonunital rings and that the statement is about proper subrings being finite, we can observe that each ideal is a subring. Also I assume $R\ne\{0\}$.
Suppose $xR$ is proper for every $x\in R$, $x\ne0$. Being a finite commutative ring with no nonzero divisors, $xR$ is a field, so it has an identity $xe\ne0$. Then $xexr=xr$, for every $r\in R$, so $exr=r\in xR$ and $xR=R$. A contradiction.
Hence, for some $x\ne0$, we have $xR=R$. Then there is $y\in R$ with $xy=x$. In particular, $xR\subseteq yR$, so also $yR=R$ and there exists $z$ with $yz=y$.
If $r\in R$, then $ry=ryz$, so $r=rz$. Hence $R$ has an identity $1=z$.
Now, let $r\in R$, $r\ne0$. Then the minimal subring $S$ of $R$ containing $1$ and $r$ is either finite or $R$. In the first case $S$ is a field, so $r$ is invertible in $S$ and hence in $R$ (they share the identity).
Assume $S=R$. Let $P$ be the prime subring of $R$; then there is a surjective homomorphism $P[X]\to R$, sending $X$ to $r$. If this homomorphism is injective, then the image of $P[X^2]$ is a proper subring of $R$ and is infinite: a contradiction. Then the homomorphism has a nontrivial kernel and so $R$ is a finite field.
