Prove that the area of their union is greater than $\frac{2}{9}S$ 
A finite set of unit circles is given in a plane such that the area of their union $U$ is $S$. Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater than $\frac{2}{9}S$.

What I am wondering is what if our finite set of circles is just like $10$ circles that are all in the same position in space. Then none of them is disjoint. How does there exist a subset of mutually disjoint circles with the desired property?
Or if that example was bad how do we know that there will always exist a subset of mutually disjoint circles? Maybe that will include only one circle?
Here is the official solution.

 A: Here's an explanation of the official solution:
Cover the plane in a tessellation of hexagons with inradius $2$. Then, put all of the hexagons on top of each other so all of the plane fits onto this one hexagon. This is kind of confusing, but basically, you have to visualize all of the circles becoming moved onto this one hexagon based on where they are in the tessellation. If they are at the top of one of the hexagons, they are now at the top of this hexagons; if they are at the bottom of one of the hexagons, they are now at the bottom of this hexagons; if they were partially in the top of one hexagon and in the bottom of another, part of them is now at the top of this hexagon and part of them is at the bottom of the hexagon.
Now, you have mushed an area of $S$ into an area of $8\sqrt 3$. This means that you must have some point that has $\frac{S}{8\sqrt 3}$ overlaps from circles that were originally from different hexagons. Understanding this takes some visualization, but basically, if $S > 8\sqrt 3$, then obviously, you will have some overlap from two different hexagons because you can mush more area into less area without overlapping. If you have $S > 16\sqrt 3$, then you have already covered the area two times, so now you need to overlap from three different hexagons in order to mush more area in. This pattern until it gets to how if you have $S$ area, you need overlap from $\frac{S}{8\sqrt 3}$ hexagons.
Once you have that figured out, now you need to visualize how if two circles overlap the same point on this hexagon and they are from different hexagons, then they are disjoint. This takes a lot of visualization and you need to account for the fact that the inradius of the hexagon is $2$ because if the inradius were smaller, this would not work. Try to draw or visualize different examples of tessellations and unit circles over those tessellations to see how this works. Make sure you draw unit circles over the border in order to deal with the border cases. Basically, the reason why unit circles can't overlap without being from different hexagons is because to overlap, they go at least the distance of the indiameter of $4$ and two unit circles can not overlap while overlapping points $4$ units apart because $4$ units is the sum of the diameter of two unit circles.
Now, we know that there is a point where there are overlaps from $\frac{S}{8\sqrt 3}$ different hexagons and that the circles from these hexagons are all disjoint. Thus, we have found $\frac{S}{8\sqrt 3}$ disjoint circles, meaning we have a set of mutually disjoint circles and since each circle has an area of $\pi$, the area of these circles altogether is $\pi\frac{S}{8\sqrt 3}=\frac{\pi}{8\sqrt 3}S>\frac 2 9 S$.
A: Take the example you gave: There are $10$ congruent circles all completely overlapping with each other. Thus, the area of any one circle is the same as $S$, which is the area of the union of all of the circles, because the union of all of the circles is the same as any one of the circles. Thus, if one takes a subset containing just one circle, then that subset is mutually disjoint because there is only one circle and the area of the union of that set is $S$, which is clearly greater than $\frac 2 9 S$.
Just keep in mind that a set of one circle is mutually disjoint.
