The planar case is rather simple but I'm still struggling with dimension 3 and higher, so we'd better keep this thread open at least for a while.
Lemma 1: Suppose we have finitely many infinite strips of total width at least $8$. Then we can move them (without rotations) so that they cover a disk of radius $1/4$ (the exact numbers do not matter; all that is essential is that they are two absolute constants).
Proof: We can choose a subset of strips with total width at least $1$ so that all directions lie in an angle of size $\pi/4$. WLOG, that angle is $0\le\theta\le \frac \pi 4$, where the angles are counted from the vertical up direction counterclockwise. Arrange the strips so that their directions satisfy $0\le \theta_1\le\theta_2\le\dots\le\theta_n\le\frac\pi 4$. Let $w_1,w_2,\dots,w_n>0$ be their widths. We will think of our strips as going "vertically", so each strip has the "left" and the "right" edge. Now position the $j$-th strip so that its right edge passes through the point $(w_1+\dots+w_j,0)$ on the horizontal axis. By induction, we can see that then the first $j$ strips cover the triangle bounded by the horizontal axis, the vertical axis, and the line passing through $(w_1+\dots+w_j,0)$ in direction $\theta_j$ (note that this triangle doesn't need to depend on $j$ in a monotone way!). Thus, in the end we'll cover the triangle bounded by the axes and the line through $(w_1+\dots+w_n,0)$ in direction $\frac \pi 4$, which, since $w_1+\dots+w_n\ge 1$, contains a disk of the required size.
WLOG all our convex sets $U_j$ are rectangles with sides $a_j\le b_j$; moreover, we can think that $b_j$ is always some negative power of $2$.
Lemma 2: Assume that we have some finite collection of $U_j$ with common $b$ and various $a$'s and orientations. If $\sum_j a_j\ge 8b$, then we can use them to cover a disk of radius $\frac b4$.
Proof: This is just Lemma 1 in disguise: scaling by a factor of $b$ does not matter and the piece of any strip of width $a_j$ in the disk so small is always contained in a rectangle $a_j\times b$ in the same direction.
Now we just take a fixed $b$ and start dividing rectangles into finite families with sum of $a_j$ between $8b$ and $9b$. Each family can be used to cover a disk of area comparable with the total area of the family and the useless remainders will have total area of $8b^2$ or less, so the sum of remainder areas will be finite.
This reduces the problem to the case when all convex sets are disks, which can be handled similarly to the classical Vitali lemma: just consider 2 times smaller disks and throw them into the unit disk $D(0,1)$ (at least their center has to be thrown in that disk), without intersection with previously thrown disks and throwing them in decreasing order of radius as long as you can. This process must terminate after a finite number $N$ of steps since at every step the yet uncovered area decreases with the area of the disk currently being thrown and those areas sum to infinity. For later purposes, recollect $0<r_{N+1}(\leq r_N\leq...)$ as the radius of the disk which was just about to be thrown before the termination of our throwing-process. Now consider the configuration of disks $\{D(x_j,2r_j)\}_{1\leq j \leq N}$ with the same center-positions $x_j$ but double the radius. We have $D(0,1)\subseteq \bigcup_{j=1}^N D(x_j,2r_j)$. For suppose that $x\in B(0,1)$ was not in that union, then $x$ is a distance of at least $2r_j \geq r_j+r_{N+1}$ removed from $x_j$. So $D(x,r_{N+1})\cap D(x_j,r_j)=\emptyset$ for all $1\leq j\leq N$. So the $N+1$'th disk can still be thrown disjointly from the previous disks, which contradicts that we had thrown disks as long as we could.
As I said, this argument is $2$-dimensional. To my shame, I do not know even if the analogue of Lemma 1 holds in higher dimensions (and, unlike the 2D case, Lemma 1 would not immediately imply Lemma 2, though the Vitali game, if one can reach it, can be played in any dimension). Any bright ideas?
