Suppose you have the following product
$\mathbb{R}^n \times \lbrace 0_{\mathbb{R}^{N-n}} \rbrace$
and another open set $V \subset \mathbb{R^N}$. Is this still an open set if we have the intersection of both sets so to say:
$V \cap \mathbb{R}^n \times \lbrace 0_{\mathbb{R}^{N-n}} \rbrace$.
If we after this take the N-n last entries (equal to 0) of the resulting vectors and cut them of, then we get a subset of $\mathbb{R}^n$. Is this open?
Yes, $\mathbb R^n$ is a subspace of $\mathbb R^N$ and subspaces inherit properties of openness.
Don't think of it as intersecting it with the $A \times \{0_{\mathbb R^{N-n}}\}$ first (why that set? Why not any other set?) Just think of it as "cutting off" the last $N-n$ terms from the very beginning.
The end result is this is just a projection of $V$ to $\mathbb R^n$. It doesn't matter what weird manipulations you did to it along the way. And projections of open sets into sub-spaces are open.
===== oops ==== misread the question === old answer below ==
No, it is not open unless it is empty. If $A \subset \mathbb R^n$ then $A \times \{0_{\mathbb R^{N-n}}\}$ is not open. And any intersection will be of that form.
For any point $p\in A \times \{0_{\mathbb R^{N-n}}\}$ and for any open ball of radius $\epsilon > 0$ centered at $p$ will have points that have non-zero values in the $n+1... N$ "positions". Thus no open ball of $p$ is completely contained in $A \times \{0_{\mathbb R^{N-n}}\}$. So no point $p\in A \times \{0_{\mathbb R^{N-n}}\}$ is an interior point.
See Why is the set $(0,1)\times \{0\}$ not open with respect to $\mathbb{R}^2$?