Packing infinitely many ellipses into a circle

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Given a circle $C$, and an infinite set $S$ of mutually disjoint ellipses which are inside and tangent to $C$, prove that there must exist a disk $D$ which lies inside $C$ but outside every ellipse.

It seems like there should be an elegant proof.

Note that if degenerate ellipses, i.e. line segments, are allowed, then the conclusion does not follow -- radial segments can be used that get arbitrarily close to the circle's center in every sector.

(This question was inspired by this one about filling the plane with parabolas.)

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This is hard to make rigorous, but if assume by contradiction that all the space inside $C$ is taken, then there are couples and triples of ellipses arbitrarily close. Assume that we have $3$ ellipses very close each other, like in figure: ellipses arrangement then they define a region $D$ with a concave boundary with the property that any entering ellipse cannot have a minor axis greater than the separation between the ellipses. Hence I would bet that the measure of the subset of $D$ taken by entering ellipses cannot exceed the sum of the areas of the three depicted trapezoids, and such a sum is stricly less than $\mu(D)$, hence in $D$ there must be a neighbourhood of a point that is not taken by any ellipse, as wanted.

However, this argument just gives an idea and a bet, and it is still far from being a proof.