2.20

As the hint prescribes, define: $$\mathcal{C} = \lbrace E \subset \mathbb{R}^d \hspace{0.25cm} | \hspace{0.25cm} E^y \in \mathcal{B}(\mathbb{R}) \rbrace$$
First, note that $\mathcal{C}$ is trivially non-empty. Next, does $E \in \mathcal{C} \hspace{0.25cm} \Rightarrow E^c \in \mathcal{C}$?

Yes. Notice $(E^c)^y = (E^y)^c$ (with respect to the slice, not $\mathbb{R}^2$). Thus, since $E^y$ is Borel, $E^c \in \mathcal{C}$.

Is $\mathcal{C}$ closed under countable unions and intersections? Yes, using identical reasoning from above. Thus, $\mathcal{C}$ is a $\sigma$-algebra.

Does $\mathcal{C}$ contain the open sets?

Indeed let $\mathcal{O}$ be an open set in $\mathbb{R}^d$. Notice that for any $x$ lying on any slice $\mathcal{O}^y$ $\exists r > 0$ such that $B_r(x) \subset \mathcal{O}$ Certainly, $B_r(x)^y \subset \mathcal{O}^y \hspace{0.25cm} \Rightarrow \mathcal{O}^y$ is open, and therefore Borel. Thus, any open set in $\mathbb{R}^2$ is in $\mathcal{C}$.

Now, since $\mathcal{B}(\mathbb{R}^2)$ is the smallest $\sigma$-algebra that contains the open sets in $\mathbb{R}^2$, we have that $\mathcal{B}(\mathbb{R}^2) \subset \mathcal{C}$. Thus, every slice of a Borel set $E$ must be Borel, as well.