Connectedness property of $R^2$

My class teacher proposed this problem which seem very interesting.

If we remove countably many open disc from $R^2$. Is the remaining space still be path connected.

I have done the problem for the case when disc are not disjoint and find out that space need not be path connected. But I am wondering what happen if disc are disjoint.my guess is space should be path connected any help will be appreciated.

The set is path-connected. Given $A$ and $B$, construct a path from $A$ to $B$ by starting with the straight line segment from $A$ to $B$, with uniform speed; whereever this is in the space, the path is given by the line segment; wherever it is not, this is because of a specific disk; define the path to map the interval that would have been mapped to a chord of this disk to the corresponding minor arc with constant speed (where if the line segment cuts the disk in half, choose, say, the left-hand arc). The arc is covered neither by this disk, nor by any other disjoint disk. Continuity follows from the fact that the remapping cannot stretch distances along the path by more than a factor of $\pi/2$.

This might be a counter example where I replace open balls by open rectangles.

let consider the set {$1/n$}... if $n$ is odd thne consider the open rectangles $(1/n,1/n+1) \times (2k, 2k+2)$ s.t $ k \in \mathbb{Z}$ ...and if $n$ is even then consider open rectangles $ (1/n,1/n+1) \times (2k+1, 2k+3)$ s.t $ k \in \mathbb{Z}$... now my claim is that if I take out all such rectangles from $\mathbb{R^2}$ , then the remaining space cannot be be path connectedto prove this we can use some similar kind of argement which we usually do while proving that topological sine curve is not path-connected...