No, such $X$ does not exist.

Assume the contrary. Take a point $O$ which is outside all disks and a circle $\omega$ centered at $O$. We implement polarity with respect to this circle (thus, in what follows we consider only lines not passing through $O$).

Take any ball $B_i\in X$. The poles of all lines tangent to $B_i$ form a hyperbola $h_i$ with a focus at $O$; then the poles of all lines intersecting $B_i$ form the set $H_i$ consisting of $h_i$ and of the interior points of its two branches. Let $A_i$ be the cone bounded by the asymptotes of $h_i$ and containing $H_i$ (we assume that $A_i$ contains no points of the asymptotes except for $O\in A_i$). Notice here that in any closed subcone of $A_i$, all points which are sufficiently far from $O$ lie in $H_i$.

The fact that $X$ satisfies the required properties yields that the sets $H_i$ cover the plane (except for $O$) but in a bounded number of layers. We will show that this is impossible.

Take an arbitrary $A_{i_1}$ and some its closed subcone $S_1$. Some interior point $x_1$ of $S_1$ is not covered by $H_{i_1}$, so it is covered by some $H_{i_2}$. Take a closed subcone $S_2$ of $S_{i_1}\cap A_{i_2}$ and proceed similarly. Thus, on the $(k+1)$th step we choose an interior point $x_k\in S_k$ not covered by $H_{i_1},\dots,H_{i_k}$ (all points close enough to $O$ are such), take $H_{i_{k+1}}$ covering $x_k$, and take a closed subcone $S_{k+1}$ of $S_k\cap A_{i_{k+1}}$.

Finally, for every $k$, all points in $S_{k}$ which are sufficiently far from $O$ are covered by each of $H_{i_1},\dots,H_{i_k}$, so some point is covered by arbitrarily many of the $H_i$.