Delete some circles to isolate each one while still cover enough area. Problem
Let $\mathcal C$ be a finite set of unit circles in the Euclidean plane such that the area of the union of the circles in $\mathcal C$ is $A$. Then there exists a subset $\mathcal C'$ of $\mathcal C$ such that:
$\bullet$ No two distinct circles in $\mathcal C'$ intersect, and
$\bullet$ The area of the union of the circles in $\mathcal C'$ is at least $2A/9$.
My Attempt
Let $G=(V, E)$ be a graph whose vertex set consists of $|\mathcal C|$ symbols, one for each member of $\mathcal C$. Two vertices $u$ and $v$ are adjacent if the corresponding circles intersect.
Note that for any vertex $v$, the area of the union of the circles corresponding to the vertices in $v\cup N(v)$, where $N(v)$ is the set of all the neighbors of $v$, is no more than $9\pi$.
Now let $S$ be a maximal independent set in $G$, and $\mathcal C'$ be the set of circles corresponding to the vertices in $S$. We claim that the area of the union of the cirlces in $\mathcal C'$ is at least $A/9$.
Let us write $C_u$ to denote the member of $\mathcal C$ corresponding to a vertex $u\in V$.
Since $S$ is a maximal independent set, every vertex in $G$ is a neighbor of some vertex in $S$.
Thus we have
$$
\sum_{v\in S}9\cdot\text{area}(C_v)
\geq
\sum_{v\in S}\text{area}\left(\bigcup_{u\in\ v\cup N(v)} C_u\right)
\geq
\text{area}\left(\bigcup_{u\in V}C_u\right) = A$$
giving
$$\sum_{v\in S}\text{area}(C_v)\geq A/9$$
Also, since $S$ is an independent set, no two circles in $\mathcal C'$ intersect.
So instead of $2A/9$, I am abnle to achieve $A/9$. How can we improve this?
 A: An improved bound
Form the convex hull of the centres of the circles and consider a circle $C$ with centre $v$ on the boundary of the convex region. Delete $C$ and all circles which intersect $C$.
The deleted area is at most the sum of the areas of a half-circle of radius $3$, two quarter circles of radius $1$ and a $1$ by $4$ rectangle i.e. $5\pi+4\approx6.27\pi$. This deleted area contains the circle $C$ which does not intersect any of the remaining circles.
Proceeding in this fashion with the reduced set of circles we see that we obtain approximately $$\frac{1}{6.27}A\approx 0.16A.$$
A: An improved bound
$\pi A/8\sqrt{3}$ (≈0.2267A), better than 2A/9 (≈0.2222A) in problem. A brief solution can be found here, problem 11.
This solution is a bit similar to Shapes in a lattice, but uses hexagonal lattice instead of square. (Square lattice can only get the bound of $\pi A/16$ (≈0.1963A).)
Let's put the shape $\mathcal C$ into a hexagonal lattice with hexagon's inradius 2 (exradius and edge of hexagon is $4/\sqrt{3}$). Then we stack all hexagons like Shapes in a lattice.
Case 1
For example, a shape $\mathcal C$ with 2 intersected circles is in red and lattice is in black:

After stacking, the shape looks like:

There is no overlap in this case, so the total area A of $\mathcal C$ is at most the hexagon $8\sqrt{3}$, i.e. $A\le 8\sqrt{3}$.
Just pick one circle as $\mathcal C'$ then we get $A'=\pi$, so $A'\ge\pi A/8\sqrt{3}$.
Case 2
For overlapped cases, for example, we just add another circle in the center of a hexagon into Case 1:

The shape after stacking looks like:

with overlap in green. Here we denote the maximum overlap count as N (N = 2 in this case; N = 1 in Case 1, which means no overlap).
So the total area A of $\mathcal C$ is at most N hexagons, i.e. $A\le 8N\sqrt{3}$, or $N\ge A/8\sqrt{3}$.
If we pick a point B in the overlap area, there must be N disjoint (that is the reason we choose hexagons of inradius 2) circles containing the point B:

In this case the area of $\mathcal C'$ is $A'=N\pi$. So we get $A'\ge\pi A/8\sqrt{3}$ again.
Ultimate bound?
I think the ultimate bound of this problem is $A/4$ (0.25A), which may not be solved by stacking lattice unit.
Case 3
Let's draw circles as many as possible with only a very small area of intersections like:

The total area of $\mathcal C$ is close to $4\pi$ but the area of $\mathcal C'$ is $\pi$.
However, I can neither prove it nor think out a counter example.
There may also be an ultimate bound in problem Shapes in a lattice, which looks unable to solve by stacking. See Shapes in a lattice: area less than 1 or π/2?.
