Actually, since your set $Y$ is itself a Borel (in fact, closed) subset of $\mathbb{R}$, it turns out at the Borel subsets of $Y$ are exactly the Borel subsets of $\mathbb{R}$ which happen to be subsets of $Y$.
In general, if $Y$ is an arbitrary subspace of some topological space $X$, the Borel subsets of $Y$ are of the form $Y \cap E$ where $E$ is a Borel subset of $X$. The reason for this is because the open subsets of $Y$ are exactly those sets of the form $Y \cap U$ where $U$ is an open subset of $X$ (though the proof is somewhat more involved). If $Y$ is a Borel subset of $X$, then $Y \cap E$ is also a Borel subset of $X$ for all Borel $E \subseteq X$, which helps gives the characterisation from above.