Problem: Decide whether it possible to cover all lattice points in $\RR^2$
by an (infinite) family of disks whose interiors are disjoint
such that the radius of each disk is at least $5$.

Solution: The answer is no.

Assume not.
Take a disk $\odot O$ not touching any member of the family,
and then enlarge it until it is maximal.
Then, it must be tangent to at least three other disks,
say $\odot A$, $\odot B$, $\odot C$.
Suppose WLOG that $\angle AOB \le 120\dg$.
Denote the radii of $\odot O$, $\odot A$, $\odot B$ by $r$, $a$, $b$.

But the Law of Cosines gives
\[ (a+b)^2 \le (a+r)^2 + (b+r)^2 + (a+r)(b+r) \]
which rewrites as
\[ 12r^2 \ge (a-3r)(b-3r) \ge (5-3r)^2 \]
which one can check is impossible for $r \le 1/\sqrt2$.
Thus $r > 1/\sqrt2$.

In particular $(\odot O)$ must contain a lattice point
as it contains a unit square.

\begin{remark*}
  The order of the argument here matters in subtle ways.
  A common approach is to try and reduce to the ``optimal'' case
  where we have three mutually tangent circles,
  and then apply the Descarte circle theorem.
  There are ways in which this approach can fail
  if the execution is not done with care.
  (In particular, one cannot simply say to reduce
  to this case, without some justification.)

  For example: it is not true that, given an infinite family of disks,
  we can enlarge disks until we get three mutually tangent ones.
  As a counterexample consider the ``square grid''
  in which a circle is centered at $(10m, 10n)$ for each $m,n \in \ZZ$
  and has radius $5$.
  Thus it is also not possible to simply pick three nearby circles
  and construct a circle tangent to all three:
  that newly constructed circle might intersect a fourth disk
  not in the picture.

  Thus, when constructing the small disk $\odot O$
  in the above solution,
  it seems easiest to start with a point not covered and grow $\odot O$
  until it is tangent to \emph{some} three circles,
  and then argue by cosine law.
  Otherwise it not easy to determine which three
  circles to start with.

  In all solutions it seems easier to prove that a disjoint circle
  of radius $1/\sqrt2$ exists,
  and then \emph{finally} deduce it has a lattice point,
  rather than trying to work the lattice point into the existence proof.
\end{remark*}