Decompose $\mathbb{R}^N = \mathbb{R}^{N_1} \times \mathbb{R}^{N_2}$, let $\Omega \subset \mathbb{R}^N$ be a bounded domain and take $x$ in the set $$\Pi_1(\Omega) = \{ x \in \mathbb{R}^{N_1} : (x,y) \in \Omega \text{ for some } y \in \mathbb{R}^{N_2}\}. $$ Define the set $$ A_x := \{y \in \mathbb{R}^{N_2} : (x,y) \in \Omega\}. $$ I am trying to discover what should be the boundary of $A_x$. My first attempt was $$ (1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\partial A_x = \{y \in \mathbb{R}^{N_2} : (x,y) \in \partial \Omega\}, $$ what, I think, is not true.
Edit: For me it is enough to prove: If $y \in \partial A_x$ then $(x,y) \in \partial \Omega$. Is it true ? Here there is a drawing which shows that (1) may not be true:
Notice that $x$ in the picture is in $\Pi_1(\Omega)$ and $y_0$ in the picture is in $\{y \in \mathbb{R}^{N_2} : (x,y) \in \partial \Omega\}$. However $y_0 \not\in \partial A_x$.
For all $x\in \mathbb R^{N_1}$, the inclusion $$ \{x\} \times \partial A_x \subset \partial \Omega\quad (*) $$ (that you are asking about in the end of the edit) is true. At the same time, as your example shows, the opposite inclusion is false even if you assume that $A$ is open with smooth boundary and $x\in \Pi_1(A)$.
Let's prove the inclusion (*). (I do not need any assumptions on $A$ and $x$.) Take $p=(x,y)$ such that $y\in \partial A_x$. Let $U=U_x\times U_y$ be a product neighborhood of $p$. Since product neighborhoods form a basis of topology at $p$, it suffices to prove that every such $U$ has nonempty intersection with both $A$ and its complement $A^c$ in $\mathbb R^N$. Since $y\in \partial A_x$, there exists $z\in A_x\cap U_y$. The point $(x,z)\in A$ lies in $U$ since $x\in U_x$. Thus, $U\cap A\ne \emptyset$. Furthermore, since $y\in \partial A_x$, there exists $w\in U_y\setminus A_x$. Then $(x,w)\notin A$ either. Hence, $U\cap A^c\ne\emptyset$ as well. Thus, $p\in \partial A$. qed
Addendum. Call a boundary point $p\in \partial A$ hidden if the equality in (*) fails. One can prove that, in a sense, $p$ can be hidden only in the situation described in your example. More precisely, assuming that $A$ is open with smooth boundary, and $p$ is a hidden boundary point then the affine subspace $\{x\}\times {\mathbb R}^{N_2}$ is contained in the tangent space $T_p\partial A$. (But the inclusion $\{x\}\times {\mathbb R}^{N_2}\subset T_p\partial A$ does not imply that $p$ is hidden.)