Do ellipsoids cast ellipsoidal shadows?

Yes they do. You can prove it by induction on the codimension of the subspace you project to. For $x\in Vect(e_1,\ldots e_{n-1})$ there exists $t \in \mathbb{R}$ such that $x+te_n$ belongs to $\Delta$ iff the discriminant of the degree $2$ equation $(x+te_n)^TA(x+te_n)\leq c$ w.r.t. the unknown $t$ is non-negative, which turns out to still be a quadratic inequality in $x$.

Yes. An ellipsoid is a linear transformation of a spherical ball, and orthogonal projection is also a linear transformation, so it suffices to show that any linear transformation whose image is a subspace sends a spherical ball to an ellipsoid in that space.

A linear transformation can be decomposed into orthogonal projection by its kernel followed by some invertible linear transformation. Orthogonal projection sends a spherical ball to a spherical ball in the subspace, so we are done.

Indeed, ellipsoids cast ellipse shape shadows on the ground.

The intersection of any conicoid and a first degree equation plane illumination terminator between two tangential points is a conic section. It can be proved by elimination to the conic second degree equation.

There are already a good answers presented, but I want to add the also one may think in a following way :

The orthogonal projection defines some subspace $\langle e_1, e_2 \ldots e_n \rangle$, and we perform an orthogonal transformation $R^{T}$, such that the matrix $A$ transforms into $R^{T} A R$, and in the rotated basis, the first $n-1$ components will correspond to that subspace. After the rotation, matrix $A$ preserves its positive definiteness, and the restriction to the $(n-1) \times (n-1)$ will be positive definite by the Sylvester's critertion. Therefore this block would define an ellipsoid in one dimension lower.