Can every $\mathbb{Z}^2$ disk be pinball-reached? Let every point of $\mathbb{Z}^2$ be surrounded by a mirrored disk of radius $r < \frac{1}{2}$,
except leave the origin $(0,0)$ unoccupied by a disk.

Q. Is it the case that every disk can be hit by a lightray emanating from the origin
  and reflecting off the mirrored disks?

Lightrays are composed of (infinitely thin) segments, and reflect off the disks
with angle of incidence equal to angle of reflection.
For example, here is one way (of many ways)
to hit the $(0,2)$ disk when $r = \frac{1}{4}$ with two reflections;
it clearly cannot be reached directly, with zero reflections:



I believe the answer to my question Q is Yes, but I would be
grateful for confirmation from the dynamical systems experts.
(Forgive me if I have not learned sufficently from my previous, related question, 
"Pinball on the infinite plane.")
It occurs to me it might be interesting to color the disks according to the minimum number
of reflections needed to hit each...
 A: I add this image just to illustrate that matters seem more complicated (as Douglas Zare has emphasized)
when the radii of the disks approaches $\frac{1}{2}$:
          
A: For small radii r, perhaps r less than 1/5, something like the following should work.  I use symmetry to
restrict my attention to circles in the first quadrant.
Use a checkerboard coloring and color the origin and circles with coordinates of like parity the same
color, e.g silver.  It should be clear that any silver circle with coordinate (0,n) or (1,n) is reachable by
using n-1 reflections, and that at least 120 degrees of arc on that circle is reachable.
Now one can cover larger x coordinates by reflecting the ray off (0,n) and going up.  Although the
ray does not emanate from the center of (0,n), it should be clear that there is enough of the or 
spectrum of angles to choose from that one can use an additional reflection to hit, say, (2,n) after
leaving (0,n).  This should generalize to an arbitrary silver circle, and each such has at least
120 degrees of arc as an available target.
Once all the silver circles are shown to be reachable, construct a path to an arbitrary circle (m,n) by traversing to
(0,n) or (0,n+1), which ever is silver, go up to a silver circle past but near (m,n), and reflect off a silver circle to the desired target.
Gerhard "Loopy After Bouncing Off Walls" Paseman, 2013.04.19
A: Permit me to draw attention to a new expository survey related to this question:

Alex Wright. "From rational billiards to dynamics on moduli spaces." Apr. 2015.
  (arXiv abstract.)

He discusses the "Wind Tree" model, also known as the Ehrenfest model,
which is essentially the model I was considering, but with square (or rectangular)
"trees" rather than disk trees:

     


A: Douglas Zare's shortest path idea seems to me very well-suited for this.
Intuitively, we can view the circles as being rings, and the reflected ray like a rope going through the rings. We pull to obtain the shortest rope (considering the rings fixed, and other suitable idealizations).
The picture below shows how a path connecting $(0,0)$ with $C(m,n)$ may be. Normally, one should be able to calculate the precise contact points and the reflection angles, from the initial angle and $r$, but I am too lazy to do this.

Added
Douglas Zare's comment, that when the radius is close to $1/2$, the things get more difficult, is right, as it can be seen from the illustration provided by Joseph O'Rourke in another answer. So here's how I think we can use the solution I presented above, to handle any possible $r<1/2$.
Start with the above solution, which works for, say, $r_0=1/3$. If $r<1/3$, one can decrease the radii of the circles, until the desired radius is reached, without aby problem. The solution will still hold.
The difficulties appear if the radius is larger. We gradually increase the radii of the circles, until the string becomes tangent to at least a circle. Then, we wrap once again the string. We continue to gradually increase the radius, and when the string becomes tangent, wrap it more, until we get the desired radius. The two main cases we can encounter with the solution I presented above, along with the proposed "moves", are represented below.

A: Let $C_r(x,y)$ or $C(x,y)$ be the circle of radius $r$ about the lattice point $(x,y)$.
Suppose we choose a sequence of circles to hit, and ask for the piecewise linear path of shortest length from the origin hitting each of the circles along the path. If this doesn't go inside a circle, then by the least action principle, the angle of incidence will equal the angle of reflection. 
We can run into problems in two ways. First, the line segments can intersect other circles. For example, if we ask the path to visit $C(1,0)$ and then $C(3,0)$, then the line segment must intersect $C(2,0)$. So, we better restrict the paths not to do that. Second, the shortest piecewise linear path may pass through the interior of a circle. For example, from the origin to $C(1,1)$ to $C(2,2)$, the shortest path passes through the interior of $C(1,1)$. Again, to avoid this, we'll restrict the paths.
If $r$ is close to $1/2$, then you will need to bounce back and forth several times between adjacent circle to squeeze by them. However, for smaller $r$, we can construct a viable path more simply. Suppose $r \lt \sqrt{2}/4 \approx 0.354$. Then no line segment connecting $C_r(x,y)$ to $C_r(x+1,y+1)$ passes through any other circle. Consequently, from any point on $C_r(x,y)$ to any point on $C_r(x+1,y+1)$, the shortest path which hits $C_r(x+1,y)$ does not go inside $C_r(x+1,y)$, it is a piecewise linear path which reflects off of $C_r(x+1,y)$.
Take a path from $(0,0)$ to $(x,y)$ with unit steps parallel to the axes so that each step is perpendicular to the previous. Without loss of generality, we can assume $0 \le x \le y$, and we can walk to $(0,1), (1,1), (1,2), ... (x,x)$. From there, we use a sawtooth pattern: $(x,x+1), (x-1,x+1),(x-1,x+2),(x,x+2),(x,x+3),(x-1,x+3)... (x,y)$. Then the shortest curve starting at the origin which hits the circles centered at these points in this order is a piecewise linear path of a light ray which reaches $C(x,y)$ by reflecting off the circles in that order.



               (Image added by J.O'Rourke)

This only handled $r \lt \sqrt{2}/4$. I believe that you can cover the case of $\sqrt{2}/4 \le r \lt 1/2$ by replacing $C(x,y) \to C(x+1,y)$ with $C(x,y)\bigg( \to C(x+1,y) \to C(x,y)\bigg)^n \to C(x+1,y)$, where the number of repetitions $n$ depends on $r$, perhaps $n=c/(1/2-r)$. 
