Covering the plane with disks
Here's a somewhat more elementary proof (no Baire property :-) that explicitly constructs a point not covered by a given set of (almost) non-overlapping disks.
By fixing an enumeration of the points with rational coordinates, we can enumerate the disks in the order in which we encounter them when enumerating these points. (If two disks touch at such a point, we can enumerate them in lexical order of their centres.) So there are countably many disks, and hence countably many pairs of disks, and hence countably many touching points of disks, and also countably many top and bottom points of disks (points with extremal $y$ coordinates).
Pick some point $p_1$ in the plane that has a $y$-coordinate different from all touching points, top points and bottom points. Proceed to cover the plane with the disks according to the enumeration until $p_1$ gets covered. Then move out of the covering disk in the positive $x$ direction to an as yet uncovered point $p_2$ just beyond its boundary with no covered points in between. This is possible, since the point you hit on the boundary is not a touching point, and only finitely many disks have been covered, so there must be a finite gap beyond the boundary. Now continue to cover disks according to the enumeration until $p_2$ gets covered, and again move out of the covering disk to an as yet uncovered point $p_3$ just beyond the boundary, but this time moving in the negative $x$ direction. Continue alternating between moving right and moving left. The result is an alternating series of moves by strictly decreasing distances (since we never move back past earlier, now covered points), which converges to some point to the right of all odd-numbered points and to the left of all even-numbered points. The limit point can't be in the interior of a disk, since the points are eventually outside each disk. But to be on the boundary of a disk, it would have to be a top or bottom point, since these are the only ones that you can alternatingly jump to the left and right of as you converge to them. But the limit point has the $y$ coordinate at which we started, and that was by assumption not the $y$ coordinate of any top or bottom point. Thus the limit point is neither in the interior nor on the boundary of any of the disks.
I tried to avoid having to make the argument about top or bottom points by using the Dirichlet test with three directions at angles of $120^\circ$ (inspired by the Apollonian gasket) instead of the alternating series test, but I couldn't make that work -- let me know if you can.
For those interested in whether a proof requires the axiom of choice: This one doesn't since we can make all choices explicit. The lexical order that decides which of two points touching at a point with rational coordinates to enumerate first is explicit; the point "just beyond the boundary" can be the first as yet uncovered point at a distance $2^{-n}$ from the boundary with no covered points in between; and we only need a single choice of an initial point and of an enumeration of the points with rational coordinates.
Assume $\mathbb{R}^2 = \cup_{n \geq 0} \bar{D}(x_n,r_n)$, and that the $D(x_n,r_n)$ are disjoint. Then $X = \cup_{n \geq 0} C(x_n,r_n) = \mathbb{R}^{2} \setminus \cup_{n \geq 0} D(x_n,r_n)$ is a closed subset of $\mathbb{R}^2$, in particular it is a complete metric space. $X$ being the countable union of closed subsets, by Baire's property there is a circle $C(x_n,r_n)$ which contains a non-empty open subset of $X$, i.e. there is a point $y \in C(x_n,r_n)$ and an $\epsilon > 0$ such that $X \cap D(y,\epsilon) \subset C(x_n,r_n)$.
Let's take a point $z \in D(y,\epsilon) \setminus \bar{D}(x_n,r_n)$ (time to draw a picture!). Then $z \in \bar{D}(x_m,r_m)$ for some $m \neq n$. So $C(x_m,r_m)$ is disjoint from $D(y,\epsilon)$, which implies that either $D(y,\epsilon) \subset \bar{D}(x_m,r_m)$, or $D(y,\epsilon) \cap \bar{D}(x_m,r_m) = \emptyset$. But $z \in D(y,\epsilon) \cap \bar{D}(x_m,r_m)$, so we get that $D(y,\epsilon) \subset \bar{D}(x_m,r_m)$, but then $D(y,\epsilon) \cap \bar{D}(x_n,r_n) \subset \bar{D}(x_n,r_n) \cap \bar{D}(x_m,r_m)$ so the latter has more than one point, which is a contradiction.