Denote by $S$ the set of closed points in $X=\mathrm{Spec}\:\mathbb{R}[x\_\alpha]$ ($\alpha \in \mathbb{Q}$) that have $\mathbb{R}$ as their residue field. There is an injective map from the set of continuous functions $\mathbb{R}\to\mathbb{R}$ to $S$. Is there a locally closed subset $Y\subset X$ such that $Y\cap S$ is equal to the image of this map?



Polynomial constraints on the values of continuous functions $\mathbb{R}\to\mathbb{R}$
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 $x\in Y$.


