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.
Addendum: the above proves only that $\bar{Y}=X$. As pointed out by OP, $Y$ is allowed to be locally closed. Let us take care of this case. Assume such $Y$ exists. By the above, $Y$ is a Zariski open of $X$, with complement $Z$, say. Take a non-constant polynomial $f\in \mathbb{R}[x_\alpha]$ vanishing on $Z$ and again choose a finite number of real values $c_\alpha$ for the appropriate coordinates such that $f(c)\neq 0$. One can complete the finite collection $c$ as to obtain a real point $x\in S$ associated to a non-continuous function. Since $f(x)\neq 0$, $x$ lies in the complement of $Z(f)$, so $x\in Y$.