Is the complement of countably many disjoint closed disks path connected?

Contrary to the request of the OP, this is still a sketch. I put it here because it points to some relevant literature, hoping that somebody more expert than me will help to clarify it. The question reminded me of this paper by Fischer and Zastrow: the relevant part is the end of p. 12. See also this answer by George Lowther.

You can assume that $\cup D_n$ is dense. Now choose some nice diffeomorphism $\phi:\mathbb{R}^2\to B_1(0)$. Once you apply it, the images of the disks are no longer disks, but $X:=\overline{B_1(0)}\setminus\bigcup_n \phi(\text{int }D_n)$ is still a planar 1-dimensional Peano continuum without local cutpoints (I'm not sure about the meaning of 1-dimensional..), so it is homeomorphic to the Sierpinski carpet by a theorem of Whyburn (Whyburn, Topological characterization of the Sierpinski curve, Fund. Math. 45 (1958) 320–324).

Our space is homeomorphic to $X\setminus \left(\partial B_1(0)\cup\bigcup_n\phi(\partial D_n)\right)$, but these sets we are removing are characterized as being the only non-separating simple closed curves. So under the homeomorphism with the Sierpinski carpet they map exactly to the boundaries of the holes (plus its exterior boundary) and we are left to show that:

The Sierpinski carpet is still path connected even after removing the boundaries of its holes and its exterior boundary.

But this is easy. Call $S'$ this new awful noncompact carpet. Put $C_1:=[\frac{1}{3},\frac{2}{3}]$, $C_2:=[\frac{1}{9},\frac{2}{9}]\cup [\frac{4}{9},\frac{5}{9}]\cup [\frac{7}{9},\frac{8}{9}]$, and so on. We have $S'=(0,1)^2\setminus\bigcup_{i=1}^\infty C_i\times C_i$.
Put also $G:=(0,1)\setminus\bigcup_{i=1}^\infty C_i$ and observe that $T:=(0,1)\times G\cup G\times (0,1)$ is path connected and $T\subset S'$.
Now fix any $p_0\in T$. Given any $x\in S'$, we wish to connect it to $p_0$ with a path in $S'$. Let $x\in Q_1$ where $Q_1$ is one of the $9$ closed cubes with side $\frac{1}{3}$ (those in which $[0,1]^2$ is divided when constructing the Sierpinski carpet). Choose some $p_1\in T\cap \text{int }Q_1$ and connect $p$ to $p_1$ with a path of length $<3$ (namely a path formed by at most three line segments).
Replace $[0,1]^2$ with $Q_1$ and iterate this step using self-similarity. Iterating infinitely many times we get a sequence $p_n\to x$ and paths in $S'$ connecting $p_n$ to $p_{n+1}$ with length $<\frac{3}{3^n}$, so that concatenating them we get a path from $p_0$ to $x$. $\blacksquare$

I think it is path-connected.

Let $S = \{ D_n \, | \, n \in \mathbb{N} \}$ be the set of closed disjoint disks under consideration. For any integer $m \ge 1$, let $N_m = \{ n \in \mathbb{N} \, | \, \mathrm{radius}(D_n) \in [1/m, 1/m-1) \}$ (with the convention $1/0 = \infty$) and $S_m = \{ D_n \, | \, n \in N_m \}$.

Let $P, Q \in \mathbb{R}^2 \backslash S$ and consider some path $\gamma_0$ between the two, say the straight line segment. For area considerations, there is a finite number of disks in $S_1$ intersecting $\gamma_0$ ; Let $J_1$ denote the reunion of these disks and $K_1$ denote the reunion of $J_1$ and $\gamma_0$. For area and compactness considerations, there is an $\epsilon_1 > 0$ such that the closed $\epsilon_1$-neighborhood $K'_1$ of $K_1$ intersects no other disks of $S_1$. For similar reasons, there is an $\delta_1 \in (0, \epsilon_1/4)$ such that the open $\delta_1$-neighborhood $J'_1$ of $J_1$ only intersects disks of $S$ that are contained in the open $\frac{\epsilon_1}{2}$-neighborhood of $J_1$ (and which have as such rather little radius). Consider the compact and path-connected set $R_1 := K'_1 \backslash J'_1$. One can perturb $\gamma_0$ to a continuous path $\gamma_1$ in $R_1$. Note that this implies that $\gamma_1$ is at a distance $\delta_1 > 0$ of any disk in $S_1$.

It is important to note why $J_1'$ was chosen this way : $\delta_1$ is small enough compared to $\epsilon_1$ to assure that $R_1 \backslash S_2$ is also path-connected. Moreover, points that have been pushed out at the first step don't have to be pushed out a lot in order to avoid disks of $S_2$, since it is sufficient (near those points) to perturb the path inside the intersection of $R_1$ and the $\frac{\epsilon_1}{2}$-neighbordhood of $J_1$. Points that have not been pushed out at the first step might be pushed a lot at the second step, but the same reasoning shows that they won't have to be perturbed a lot from then on.

This suggests that a by careful induction process, we can construct a nested sequence of compact path-connected sets $R_1 \supset R_2 \supset R_3 \supset \dots$ and a sequence of paths $\gamma_m \subset R_m$ joining $P$ and $Q$ such that $d(\gamma_j, S_m) > \delta_m > 0$ whenever $j \ge m$. If the perturbation process is done right (that is, if it is kind of minimal), the sequence $\gamma_m$ is equicontinuous. By the Arzelà-Ascoli theorem, we deduce that the set $\{ \gamma_m \, | \, m \in \mathbb{N}\} \subset C(I, \mathbb{R}^2)$ is relatively compact. Hence, there exists a continuous path $\gamma : I \to \mathbb{R}^2$ (still joining $P$ and $Q$) which is arbitrary well approximated by the set of $\gamma_m$. In fact, since the $R_m$s form a decreasing sequence of nested compact sets, we deduce that $\gamma \subset \cap_{m \in \mathbb{N}} R_m$. By construction, we see that $d(\gamma, S_m) > \delta_m > 0$ and consequently, $\gamma$ intersects none of the disks $D_n$.