I cannot give a full proof, but I can reduce the problem to another one that I think some people might know the answer to.
Here is the problem:
Let $U$ be an open subset of the unit square with rectifiable boundary.
$$ where $P$ is the length of the boundary of $U$ and $A$ is the area of $U$.
"PROOF:" I can prove this if $A\le \pi/4$ and I have a precise idea how to prove it in general.
First assume $A\le\pi/4$, say $A=\pi r^2$ for some $r\le\frac12$.
Since the circle minimizes the boundary length for a given area, we get $P\ge 2\pi r$ which gives the claim.
(Note that this also works if you assume that the diameter of $U$ is $\le 1$ and so this gives a second proof of Proposition 7.1 in the mentioned thesis.)
If $A>\pi/4$ this proof doesn't work. We have to replace the circle with the corresponding minimizer inside the square.
I do not know what this minimizer looks like, somewhat like a balloon you blow up inside a box. But minimal surface people might know and once you know the shape of that set, you can compute the boundary length to solve the problem as above.
Next I show how to solve the original problem once you have Problem 1.
I use induction on the number of squares.
For $n=1$ there is nothing to show.
We do $n\to n+1$.
So let a set $F$ in the plane be given which is the union of $n$ unit squares.
Let $A(F)$ be its area and $P(F)$ its boundary length. By induction hyothesis we have $P(F)\le 4A(F)$.
Add another unit square $S$, then the boundary length of the new set will be $P(F)+4-P$, where $P$ is the boundary lenght of some open set $U$ inside $S$, more precisely, $U$ is the intersection of the interior of $S$ and the interior of $F$.
Anyway, $U$ is the interior of a polygon, hence has rectifiable boundary.
The area of $F\cup S$ is $A(F)+1-A$, where $A$ is the area of $U$.
Problem 1 now tells us $P\ge 4A$.
Together with the induction hypothesis this gives the claim.