If a collection of disjoint disks covers the unit square, then the circumferences add up to infinitude. A question from Makarov & Podykorytov, Real analysis: Measures, Integrals and Applications (Can't recall what page though, but it's in the chapter about product measure).

Assume a collection of disjoint disks cover the unit square, $[0,1]^2$ up to a (Lebesgue) null set.
  Then, the sum of the lengths of their boundaries is infinitude.

My attempt:
We denote the disks $\{D_n\}_{n=1}^\infty $, their corresponding radii with $r_n$, the union $\bigcup_{n=1}^\infty D_n = C$ and Lebesgue measure on $\mathbb{R}^2$  with $m^2$,


*

*$\sum_{n=1}^\infty m^2(D_n) =\sum_{n=1}^\infty \pi r_n^2 =1 $ But this does not (and i think, cannot) produce a good bound on the sum of lengths of circumferences.

*Almost all (vertical and horizontal) cross-sections must have 1-dimensional measure $1$. I speculate this implies that up to a null-set, every cross-section intersects infinitely many disks, but did not manage to show this.

*If the last remark is true, then maybe we can show that almost-all cross-sections of the circumferences have positive measure, (as the fact that the cross-section intersects infintely many disks is encouraging).

 A: Proof 1 This proof heavily uses a result in geometric measure theory, which is not really necessary here, but is really quick and was the first that came to mind.
The result follows from Theorem 4.17 about the structure of Caccioppoli partitions in "Ambrosio, Fusco, Pallara - Functions of bounded variations and free discontinuity problems". Adapted to a partition with regular sets it states:

Theorem Given a countable partition (up to a negligible set) $C=\bigcup_n D_n$ where $D_n$ are regular and $\sum_n P(D_n)<\infty$, $\mathcal{H}^1$-a.e. point of $C$ is contained in
  $$\left(\bigcup_n D_n\right)\cup\left(\bigcup_{m\neq n}(\partial D_n\cap \partial D_m)\right)$$
  where $\mathcal{H}^1$ is the $1$-dimensional Hausdorff measure. 

In particular, $\mathcal{H}^1$-a.e. point of $\Gamma=\bigcup_n \partial D_n$ is contained in the second union.
In our case however the second union is clearly a countable set (since every pair of circumferences share at most one point) and therefore $\mathcal{H}^1$-negligible, while $\Gamma$ has positive $\mathcal{H}^1$ measure. Therefore supposing that $\sum P(D_n)<\infty $ holds for the disks and applying the theorem we obtain a contradiction (in fact the argument works for every partition with e.g. strictly convex sets).
Proof 2 This follows your proposed approach.
As you already said, almost every vertical cross-section has measure $1$. Among these, suppose that a cross-section $v=\{x\}\times [0,1]$ intersects finitely many disks. In particular, $\{v\cap D_n\}_n$ gives a finite collection of disjoint intervals which has full measure inside $v$. It follows that $v\cap \Gamma$ is made of a finite number of points, which must belong to the boundary of two disks simultaneously, but we already noticed that this set is countable. Therefore a.e. cross-section intersects an infinite number of disks. From here we can conclude in different ways:
2.1 From the Area formula for retifiable sets, another tool from GMT (see Theorem 2.91 in the same reference): calling $\pi$ the projection on the $x$ axis
$$\mathcal{H}^1(\Gamma)\geq \int_\Gamma J_1 (d^\Gamma\pi(z))d\mathcal{H}^1(z)=\int_0^1 \#\{\Gamma\cap \pi^{-1}(\{x\})\}dx=+\infty.$$
2.2 From the Crofton formula, repeating the same argument above to obtain that a.e. cross-section in any possible direction intersects an infinite number of (boundaries of) disks. Note that this is basically the area formula in disguise, in a simpler setting and with a simpler proof.
2.3 Knowing this exercise comes from the section on product measures, we can conclude in a more elementary and pertinent way: observe first that $P(D_n)=\pi \mathcal{H}^1(d_n)$ where $d_n$ is the horizontal diameter of the disk $D_n$, and call $E=\bigcup_n d_n$. Now consider the measure $\mathcal{H}^1\otimes \mathcal{H}^0=\mathcal{L}^1\otimes \mu$ where $\mu$ is the counting measure. Then by Fubini
\begin{align}\frac1\pi\sum_n P(E_n)=\sum_n \mathcal{L}^1(d_n)&=\int\limits_0^1 d\mu(y)\int\limits_0^1\chi_E(x,y) d\mathcal{L}^1(x) \\
&= \int\limits_0^1 d\mathcal{L}^1(x) \int\limits_0^1 \chi_E(x,y) d\mu(y) \\
&=\int\limits_0^1 d\mathcal{L}^1(x) \#(E\cap (\{x\}\times [0,1]))=+\infty
\end{align}
because any vertical cross-section intersecting $D_n$ will also intersect $d_n$.
