Assume that $E$ is Borel subset of $\mathbb{R}^2$. Show that for every $y \in\mathbb{R}$, the slice $E^y = \{x\in \mathbb{R} : (x, y) \in E$ is a Borel subset of $\mathbb{R}$.
Is the following enough to show that the projection of a borel set is borel or I feel that I'm missing some thing:
WLOG, let the rectangle $E=[a,b]^2$ where $a,b \in \mathbb{R}$. we can write any any rectangle as the union of closed sets as $$E=\bigcup_{n=1}^\infty [a+\frac{1}{n}, b-\frac{1}{n}]^2$$
by def above we have that for a fixed $y$
$$E^y=\bigcup_{n=1}^\infty [a+\frac{1}{n}, b-\frac{1}{n}]$$
which is indeed a borel set.
If $E=[a,b]^2$ then $E$ is a square , and $E^y=[a,b]$. But in general the projection ("slice") of a Borel set in $\Bbb R^2$ need not even be a Lebesgue set in $\Bbb R.$ See the 3rd & 4th paragraphs of https://almostsure.wordpress.com/2018/12/24/analytic-sets/