Are there countable index subrings of the reals? 
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*Does ${\mathbb R}$ have proper, countable index subrings?  By countable I mean finite or countably infinite.  By subring I mean any additive subgroup which is closed under multiplication (I don't care if it contains $1$.)  By index, I mean index as an additive subgroup.

*Given some real number $x$, when is it possible to find a countable index subring of ${\mathbb R}$ which does not contain $x$?  
 A: Perhaps surprisingly, it turns out that such subrings do exist. This was proved in Section 2 of my paper:
Simon Thomas, Infinite products of finite simple groups II,
J. Group Theory 2 (1999), 401--434. 
The basic idea of the proof is quite simple. Clearly the ring of $p$-adic integers has countable index in the field of $p$-adic numbers. Now the $p$-adic integers are the valuation ring of the obvious valuation on the field of $p$-adic numbers ... and it turns out to be enough to show that $\mathbb{C}$ has an analogous valuation. This is true because $\mathbb{C}$ is isomorphic to the field of Puiseux series over the algebraic closure of $\mathbb{Q}$, which has an appropriate valuation.
A: Simon Thomas's approach answers Question 2 too.  The answer is that for any nonzero real number $x$ there exists such a subring (possibly without 1) not containing $x$.
Proof: Let $K$ be the Puiseux series field $\overline{\mathbf{Q}}((t^{\mathbf{Q}}))$, let $A$ be its valuation ring, and let $\mathfrak{m}$ be its maximal ideal.
If $x \in \mathbf{R}$ is not algebraic over $\mathbf{Q}$, then choose an identification $\mathbf{C} \simeq K$ sending $x$ to the transcendental element $1/t$, and use the subring $\mathbf{R} \cap A$.
If $x \in \mathbf{R}^\times$ is algebraic over $\mathbf{Q}$, then choose an identification $\mathbf{C} \simeq K$ again, and use $\mathbf{R} \cap \mathfrak{m}$ (a subring of $\mathbf{R}$ without $1$). $\square$
Remark: If one insists on using subrings with $1$, then the answer is that such a subring not containing $x$ exists if and only if $x \notin \mathbf{Z}$.  
Proof: Repeat the argument above, but in the case where $x$ is algebraic (and outside $\mathbf{Z}$), use $\mathbf{R} \cap (\mathbf{Z} + \mathfrak{m})$.
A: I think the answer is no, but I didn't get anywhere. [Edit: I used to think the answer was no, but Simon Thomas convinced me otherwise.] Here is the condensed version of what I posted earlier, which seems to put serious constraints on $S$.
Let $R$ be any field and let $S$ be an additive subgroup of $R$ which is closed under multiplication. If $S$ has index less than $|S|$ as an additive subgroup of $R$, then every element of $R$ is of the form $a/b$ for $a, b \in S$. To see this, pick $r \in R$ and consider the multiples $ur$ for $u \in S$. Since $S$ has index less than $|S|$, there must be $u \neq v$ such that $a = ur - vr \in S$ then $r = a/b$ where $b = u - v$.
