= Solution
An <ellipse> here is a <nondegenerate ellipse>, hence a smooth embedded circle. Write the first <ellipse> as
$$
\gamma(t)=c+A\begin{pmatrix}\cos t\\\sin t\end{pmatrix},\qquad t\in\mathbb R/(2\pi\mathbb Z),
$$
where $A$ is invertible. Write the second as $Q(x)=0$, where
$$
Q(x)=(x-d)^TB(x-d)-1
$$
for a positive-definite symmetric matrix $B$. Its gradient is nonzero on the <ellipse>. Let $g(t)=Q(\gamma(t))$. At an intersection, <transversality> says the tangent vector $\gamma'(t)$ is not tangent to $Q=0$, so
$$
g'(t)=\nabla Q(\gamma(t))\cdot\gamma'(t)\ne0.
$$
Every intersection is thus a simple zero of $g$, and $g$ changes sign there. The intersection set is finite: it is closed and compact, and <transversality> makes each of its points isolated. Start at a parameter with $g\ne0$ and traverse one full period. Returning to the same sign requires an even number of sign changes. Therefore \b[the number of intersection points is even], including zero when the <ellipses> are disjoint.
For the upper bound, substituting the parametrization into the quadratic gives the <trigonometric polynomial>
$$
g(t)=a_0+a_1\cos t+b_1\sin t+a_2\cos2t+b_2\sin2t.
$$
It is not identically zero: otherwise the first <ellipse> would lie on the second, and its tangent would be tangent to the second everywhere, violating <transversality>. Put $z=e^{it}$. Multiplying by $z^2$ turns this function into the <polynomial>
$$
P(z)=\frac{a_2-ib_2}{2}z^4+\frac{a_1-ib_1}{2}z^3+a_0z^2+\frac{a_1+ib_1}{2}z+\frac{a_2+ib_2}{2}.
$$
It is nonzero and has degree at most four, so it has at most four distinct roots. Distinct points on the first <ellipse> correspond to distinct parameters modulo $2\pi$ and hence distinct points $z$ on the unit circle. Thus
$$
\boxed{|E_1\cap E_2|\in\{0,2,4\}.}
$$
Both the parity and the upper bound follow without using an algebraic intersection theorem. The four-point bound is attained by a unit circle and a concentric <ellipse> whose semiaxes lie on opposite sides of one.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-26-transverse-ellipses.png]
{title=Transverse ellipses with zero, two and four intersection points}
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