It seems to me that the answer is no. Call $I$ the image of this map: I claim that $I$ is Zariski dense. Since $I\subsetneq S$, this is enough to conclude. Let $f\in \mathbb{R}[x_\alpha]$ be a polynomial which vanishes on $I$. If $f$ is not identically zero then it is a non-constant polynomial in finitely many indeterminates $x_\alpha$. Choose real values $c_\alpha$ of those indeterminates such that $f(c)\neq 0$. Since this is a finite collection you can obviously find a continuous function with value $c_\alpha$ at $\alpha$, so we found a point in $I$ where $f$ doesn't vanish, contradicting the assumption.