Consider a surface S in three dimensions, such that for any polynomial function P there exists a real Z so that the intersection of the plane z=Z and S is P.; also every curve obtsined by intersection of S with a plane parelell to xy plane is a polynomial.
What is the equation/description of S
For each $n \in \mathbb{N}$, let $\alpha_n : [n-1, n] \to [-n, n]^n$ be a continuous space-filling curve that fills an $n$-dimensional cube of side length $2n$, starting and ending at its midpoint.
Combine these $\alpha_n$ into a single continuous surjective curve $\alpha : \mathbb{R} \to \oplus_{i=0}^\infty\mathbb{R}$ by setting
$$\alpha(z) = \begin{cases}0 &\text{if}& z \leq 0 \\ \alpha_n(z) &\text{if}& z \in [n-1, n], n \in \mathbb{N}.\end{cases}$$
Now let $P : \oplus_{i=0}^\infty\mathbb{R} \to \mathbb{R}[X]$ be the linear bijection defined by $P(a) = \sum_{i=0}^\infty{a_iX^i}$.
We see that the map $\Phi := P \circ \alpha$ is a surjective continuous curve over the space of all real polynomials.
Now define $S = \{(x, \Phi(z)(x), z) : (x, z) \in \mathbb{R}^2\} \subset \mathbb{R}^3$. This set is topologically a surface, since it is the graph of a continuous function $\Phi(.)(.) : \mathbb{R}^2 \to \mathbb{R}$, and it satisfies the properties you have given, namely that each horizontal slice is the graph of the polynomial function $\Phi(z)$, and since $\Phi$ is surjective, any polynomial function has a corresponding slice in $S$.
Note also that $S$ is rather crooked and definitely not differentiable, at least for $z > 0$.