If $x\in R$ is non-zero, the elements $1, x, x^2, ..., x^n$ cannot all be different, where $n$ denotes the cardinality of $R$. So $x^i=x^j$ for some $0\leq i<j\leq n$, i.e. $x^i(1-x^{j-i})=0$, hence $x^{j-i}=1$ since D is a division ring. Therefore, $x^{j-i-1}$ is the inverse of $x$ in $D$. But this element is in $R$.
Note that $R$ is automatically a unitary subring when $n>1$ (consider $x, x^2, ..., x^{n+1}$ for any $0\neq x\in R$).
Yes, assuming $R$ is a unitary subring.
If $x\in R$ is non-zero, the elements $1, x, x^2, ..., x^n$ cannot all be different, where $n$ denotes the cardinality of $R$. So $x^i=x^j$ for some $0\leq i<j\leq n$, i.e. $x^i(1-x^{j-i})=0$, hence $x^{j-i}=1$ since D is a division ring. Therefore, $x^{j-i-1}$ is the inverse of $x$ in $D$. But this element is in $R$.
Note that $R$ is automatically a unitary subring when $n>1$ (consider $x, x^2, ..., x^{n+1}$ for any $0\neq x\in R$).