Let $y \in \mathbb{R}$. If $A$ is a Borel subset of $\mathbb{R}^2$, then $$A(y) = \{x \in \mathbb{R}| (x,y) \in A\}$$ is a Borel subset of $\mathbb{R}$.
I think that I have to show that $A(y)$ is in Borel-sigma algebra in $\mathbb{R}$. But the sigma algebra itself has no clear form, I mean I just know that it is the sigma algebra generated by all open sets, there is no clear form of its elements. Another thing is I am not sure if Borel "subset" means the same as Borel 'set' (which simply mean it is in the Borel sigma algebra). Any idea or hint about how to begin ? It is very good if anyone can list some important steps in proving this statement.
Borel subset does mean the same as Borel set.
You can prove the statement by considering the set $M = \{ A\subseteq \mathbb{R}^2\,|\, \forall y\in\mathbb{R} \ A(y) \ \text{is Borel}\}$ and showing that $M$ contains all open sets and $M$ is a $\sigma$-algebra. Then $M$ must contain all Borel sets, because the Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing all open sets.${}$