First, let us consider the case in which our division $D$ of the square has no gaps. That is, the boundary between any two polygons of $D$ is either a point or a connected polygonal line.
Consider the polygon $P_{0}$ containing the center $O$ of the square ($O$ may lie on a boundary, in which case we select any of the polygons sharing that boundary). Note that $P_{0}$ lies inside a circle of radius $\frac{1}{10}$ centered at $O$. We say that $P_{0}$ is a polygon of type 1.
Now, notice that all the polygons adjacent to $P_{0}$ are inside a circle of radius $\frac{2}{10}$ centered at $O$. We call these polygons of type 2. Continuing, all the polygons adjacent to a polygon of type 2 will be called polygons of type 3, and all the polygons adjacent to a polygon of type 3 will be called of type 4, and so on.
We observe the following properties of classifying polygons by type:
- Polygons of type 5 always exist. Indeed, it is clear that $P_{0}$ and all the polygons of types 2, 3 and 4 are contained inside a circle of radius $\frac{4}{10}$. Hence, polygons of the first four types never touch the boundary of the square and we will always need to have polygons of type 5 in order for our division to fill-up the square.
- For $n\ge2$, every polygon of type $n$ has at least one adjacent polygon of type $n-1$.
- If a polygon $P$ has type $n\geq 1$, then no adjacent polygon can have type $n-1$, for otherwise $P$ itself would have type $n-1$. Hence, any two adjacent polygons must have either the same type or types differing by one.
- If a polygon $P$ of type $n\geq 2$ has less than two adjacent polygons of type $n$, then it has no adjacent polygons of type $n+1$. Indeed, for otherwise $P$ would share part of its boundary with a polygon of type $n+1$, and share another part of its boundary with a polygon of type $n-1$ (by Property 2). In lieu of violating Property 3, these two polygons could not touch. Therefore, there would be at least two regions where $P$ lies in contact with polygons of type $n$, other types prohibited by Property 3. But we are in the case that two adjacent polygons touch only along a connected polygonal segment (or a point), hence, $P$ is necessarily adjacent to at least two distinct polygons of type $n$.
The idea now is to assume that every polygon is surrounded by at most five polygons and conclude that no polygon of type 4 can be adjacent to polygons of type 5, i.e. there are no polygons of type 5, which is a contradiction by Property 1. Indded, for a polygon $P$ of type 4 we consider two cases:
If $P$ is adjacent to at most one polygon of type 4, then by Property 4 it clearly cannot be adjacent to a polygon of type 5.
Suppose that $P$ is adjacent to at least two polygons of type 4. First, $P$ will be adjacent to at least two polygons of type 3. Indeed, suppose not and let $P'$ be the only polygon of type 3 adjacent to $P$. Then $P'$ is adjacent to at least two polygons of type 4, apart from $P$- namely, the two polygons $Q$ and $Q'$ which border $P$ at the ends of its common boundary with $P'$ (see picture below)
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$Q$ and $Q'$ cannot be of type 3 since we assumed $P$ is adjacent to only one polygon of type 3 (the polygon $P'$), and they cannot be of type 5 by Property 3. Furthermore, $P'$ is adjacent to at least two polygons of type 3 (by Property 4) and to at least one polygon of type 2 (by Property 2), a total of six polygons, which is contrary to our hypothesis. Hence, $P$ is adjacent to at least two polygons of type 3, as desired.
Next, we will show that $P$ is adjacent to at least three polygons of type 3: If $P'$ is any polygon of type 3 adjacent to $P$ (we know there are at least two of these) it will not have adjacent polygons of type 4 apart from $P$. Indeed, by property 4, $P'$ is adjacent to at least two polygons of type 3. Moreover, it is adjacent to at least two polygons of type 2 (which can be proved in exactly the same way we proved that $P$ was adjacent to at least two polygons of type 3). However, since by hypothesis $P$ is adjacent to at most five polygons, $P$ must be the only polygon of type $4$ adjacent to $P'$. Now, let $Q$ and $Q'$ be the polygons touching $P$ at the ends of the boundary it shares with $P'$. These must be of type 3 and adjacent to $P$. Therefore, $P$ is adjacent to at least three polygons $P'$, $Q$, and $Q'$ of type 3, as desired.
Finally, notice that $P$ cannot be adjacent to polygons of type 5, for otherwise it would be adjacent to at least two polygons of type 4 (by Property 4), at least three polygons of type 3 (as we have just seen), which makes up at total of at least six, contrary to our hypothesis. The result is proven.
What happens when our original division $D$ has gaps? Well, as it was explained a little in the question, if two polygons $P$ and $P'$ of $D$ have gaps then the boundary they share is disconnected and these gaps are made up of polygons in our division (see the first picture in the question). We will produce a new division as follows: Add to $P$ all the gaps between it and its adjacent polygons in order to obtain a larger polygon $P_{1}$. Then do the same with all the adjacent polygons of $P_{1}$, and so on. The new division $D'$ thus formed has the property that the boundary shared by any two polygons is either a point or a connected polygonal line. Moreover, this process only decreases the number of polygons adjacent to a polygon. Hence, if some polygon of $D'$ is adjacent to at least six polygons, the same must have been true of the original division $D$. Note that each polygon $P'$ of $D'$ is contained within a polygon $Q$ obtained by adding to $P$ all the polygons adjacent to it and all the gaps between it and the polygons adjacent to it. Since by hypothesis both $P$ and all the polygons adjacent to it have diameter at most $\frac{1}{30}$, $Q$ has diameter at most $\frac{3}{10}=\frac{1}{10}$ (see picture, where the diameter of $Q$ is depicted via red lines, each of which must have length $\leq\frac{1}{30}$) and $P'$ has diameter no greater, which is what we want.
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