Covering the plane by squares!

Solved by CMU Maths Lunch Group. Misha Lavrov was the one who came up with the idea. Basically, the covering strategy is to generalize the covering strategy in $1$-dimensional case.

Lemma Given a sequence of decreasing numbers $a_1, a_2, \dots$ such that $\sum a_i^2 = \infty$. Let $n(i)$ be defined inductively: it is the smallest natural number $m$ such that $$a_{n(1) + \dots + n(i-1)+1} + \dots + a_{n(1) + \dots + n(i-1)+m} \ge 1.$$ If $b_i = a_{n(1)+\dots+n(i)}$, then $\sum b_i = \infty$.

Proof of Lemma Note that $$a_{n(1) + \dots + n(i-1)+1} + \dots + a_{n(1) + \dots + n(i-1)+n(i)-1} < 1$$ and so $$\begin{eqnarray}& & a^2_{n(1) + \dots + n(i-1)+1} + \dots + a^2_{n(1) + \dots + n(i-1)+n(i)-1} + a^2_{n(1) + \dots + n(i-1)+n(i)} \\ &\leq& b_{i-1}(a_{n(1) + \dots + n(i-1)+1} + \dots + a_{n(1) + \dots + n(i-1)+n(i)-1}) + b_i^2 < b_{i-1} + b_i^2.\end{eqnarray}$$ Summing above inequality over all $i$, we have $$\infty = \sum a_i^2 \leq \sum b_{i-1}+b_i^2 < (a_1 + 1)\sum b_i.$$

QED

We may assume the squares have decreasing length of sides $a_1, a_2, \dots$ and by compactness argument, we only need to cover the unit square. We can pick first $n(1)$ squares to cover a 1 by $b_1$ rectangle and the next $n(2)$ squares to cover a 1 by $b_2$ squares, etc. Lemma says $\sum b_i > 1$, and so we can cover the unit square.

Follows immediately from the fact that any finite collection of squares with total area $\geq 4$ can cover a square of area 1. (The proof in the link is a little complicated, but the idea is not. Shrink each square so its size is one of 1, 1/2, 1/4, 1/8, .... Then repeatedly find and combine four squares of the same area to make a larger square.)