Here, the notation $B(x,r)$ means the ball with centre $x$ and radius $r$. So in $\mathbb{R}$ this represents an interval, in $\mathbb{R}^2,$ this represents a disc, etc. Whether the ball is open or closed shouldn't matter for this question, as the boundary of a ball doesn't affect its measure (length or area).
Motivation
In $\mathbb{R}$, every open set can be written as an at most countable union of disjoint open intervals.
However, it is NOT true that in $\mathbb{R}^n$, every open set can be written as an at most countable disjoint union of open balls. For example, one can show that $\mathbb{R}^2 \setminus\lbrace{(0, 0)\rbrace} $ is not a countable disjoint union of open balls.
Now if we let $A\subset\mathbb{R}$ be an open, (possibly unbounded) set with finite, nonzero length (I believe measure is the technical term).
Then, $A$ is the union of at most countably many disjoint intervals: $ A= \displaystyle\bigcup_{i=1}^{\infty} B(y_i,r_i).$ We have:$\text{ length(} A)= \displaystyle\sum_{i=1}^{\infty} \text{ length(} B(y_i,r_i)). $ Since the length is finite, this sum converges. It is not difficult to show that there exists $k\in\mathbb{N}$ such that $\displaystyle\sum_{i=1}^{k} \text{ length(} B(y_i,r_i)) > \frac{4}{5}\text{ length(} A).\ $ Then we can find very small $\varepsilon>0,$ and finitely many points $x_i\in\mathbb{R}^2,\ 1\leq i\leq n$ where $n\in\mathbb{N},$ such that the union of the disjoint intervals is a subset of $A$, that is,
$$ \bigcup_{i=1}^{n} B(x_i,\varepsilon) \subset A $$
and has length greater than half the length of $A.$
Problem Statement
Let $A\subset\mathbb{R}^2$ be an open (possibly unbounded) set with finite, nonzero area.
Does there exist $\varepsilon>0$ and finitely many points $x_i\in\mathbb{R}^2,\ 1\leq i\leq n$ where $n\in\mathbb{N},$ such that the union of the disjoint discs is a subset of $A$, that is,
$$ \bigcup_{i=1}^{n} B(x_i,\varepsilon) \subset A $$
and has area greater than half the area of $A?$
I think this is either true by looking at $A\cup \mathbb{Q},$ or there is some fractal-like counter-example.
Note that the packing of circles in the obvious way (hexagonal lattice) has density $\frac{\pi}{2\sqrt{3}} \approx 0.9069.$
Any open set of $\mathbb{R}^n$ is an union of countably many almost disjoint cubes. (see pag 7 of http://assets.press.princeton.edu/chapters/s8008.pdf). As shown here this implies that any open set $G$ of $\mathbb{R}^n$ with finite Lebesgue measure (i.e. Area) can be approximated by a finite number of almost disjoint cubes $C_i$ such that $$ m(G / \bigcup_i^N C_i)< \epsilon $$ These cubes can be chosen to have all the same side length $l$. So it is enough to show that the theorem holds for squares of $\mathbb{R}^2$. Now consider the iscribed circle $D_i$ of each square. Its area is given by $A=\frac{\pi}{4} l^2$ So you have that $$ m(G) \ge m(\bigcup_i^N D_i)=\frac{\pi}{4} m(\bigcup_i^N C_i) \ge \frac{\pi}{4}\left(m(G)- \epsilon \right) $$
By choosing $\epsilon$ small enough you can make the area bigger than half of the area of $G$