ra.rings and algebras - Are there countable index subrings of the reals?

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 $mathbf{R} cap (mathbf{Z} + mathfrak{m})$.