There is a no set which every line meets the ball Does there exist the set of balls $X=\{B_i\subset\mathbb{R^2};i\in I\}$, satisfing following properties?(Note that the ball has a positive real radius) 


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*Let the set of all lines in plane to be $L$.  For each  $l \in L,\ l\cap B_i \ne \varnothing $ for some $i\in I$.

*$\{N_l=\text{the number of balls which the line} \ l \ \ \text{transversals}\mid l\in L\}$  is bounded. 


(This does not mean the number of ball which line transversals is finite. This statement is stronger.)  
I think there is no such set but I can not prove.
My try is think the line in the plane as a point on the sphere but I can not go further.   
Does anyone has any idea to prove or disprove?
 A: This is just a stream of thoughts, for now. Assume that such a set of disjoint, closed balls exist.
How to build it? If we start placing disks on a regular curve that disconnects $\mathbb{R}^2$, like JMoravitz suggested, its curvature must be bounded away from zero, otherwise by considering the set of tangents to the curve we have that $N_l$ is unbounded. However, a regular curve having curvature bounded away from zero either is closed, then it is enclosed in a single disk, or intersects a line at infinitely many points (like a spiral), so that approach does not work.
So, let us try to place such balls along concentric circles. We may place $3$ balls on the unit circle, $2\cdot 3$ balls on the circle with radius $3$, $2^2\cdot 3$ balls on the circle with radius $3^2$ and so on, keeping the radius of the balls always the same, $\frac{1}{2}$, and placing them accordingly to the construction of a Cantor set in $S^1$:

In such a way, a line $l$ through the origin (as well as a line that gets sufficiently close to the origin) is captured almost surely by our net, and $N_l$ is bounded by some absolute constant since the series $\sum_{n\geq 1}\frac{2^n}{3^n}$ is converging. But how to deal with lines with a big distance from the origin?
Then I wondered: maybe we can recover JMoravitz' argument by dropping the regularity condition. So we take two equilateral hyperbolas, then we take many circular arcs on these hyperbolas and place our balls there:

With an extra ball in the origin to capture the asymptotes, this configuration might work, maybe adjusting the radii of the balls as the distance from the origin increases, in order to avoid that a line $l$ parallel to an asymptote has an unbounded $N_l$.

Then a possible path for a disproof. Assume that there is a set $S$ of disjoint closed disks in $\mathbb{R}^2$ and a constant $C\in\mathbb{Z}^+$ such that every line in the plane intersects at least one disks but no more than $C$ disks. For every $R>0$, there are a finite number of elements of $S$ inside the region $x^2+y^2\leq R^2$. Otherwise, $S$ has an accumulation point $p$ in the region, and some line through $p$ has an unbounded $N_l$ (this is a delicate part ). That implies that $S$ is a countable set $^{(*)}$, so every line in the plane can be associated with a non-empty subset of $\mathbb{N}$ with at most $C$ elements, that represents the disks met. At least strange, since the lines are an uncountable set while $\mathbb{N}^C$ is a countable set, and every subset of $S$ cannot represent any line outside its convex envelope. If we map any line that does not go through the origin into an element of $\mathbb{R}^+\times S^1$, we may define the measure of a set of lines as the Lebesgue measure of the represented set in $\mathbb{R}^+\times S^1$, then compute the measure of the set of lines represented by any subset of $S$ with cardinality $\leq C$.
$^{(*)}$: in particular, by projecting the elements of $S=\{s_1,s_2,\ldots\}$ 
 on the $x$-axis, we have that $\sum_{n\geq 1} r_n$, where $r_n$ is the radius of $s_n$, must be a divergent series. Otherwise, there is some vertical line that cannot be captured by $S$.
With the previous line-point identification, it is useful to figure what are the lines captured by a single disk:

so assuming that the distance of the centre of $s_n$ from the origin is $d_n$ and $d_n\gg r_n$, the measure of the captured lines is about $2\pi\, d_n r_n$ (by approximating the epitrochoid with an annulus). It is interesting to consider the intersection between the shaded region and the unit disk having centre in the origin: since every line has to be captured by no more than $C$ disks, that sectors in the unit disk cannot overlap too much. So we have that $$\sum_{n\geq 1}\frac{r_n}{d_n}$$ is a convergent series, and:
$$\prod_{n\geq 1}\left(1-\frac{r_n}{d_n}\right)$$
is a non-zero convergent product. If we go back to the "Cantor net" configuration, that gives that a line through the origin has a positive probability to escape every disk placed along a circle centered in the origin. Hence, in order to capture these "fugitive lines", some element of $S$ must enclose the origin. And now, boom: the origin is an arbitrary point, so our set $S$ has to be dense in $\mathbb{R}^2$. Dense and closed. So, finally: 

$S$ does not exist.

