How many units squares can fit on $\mathbb R^2$?

Question

How many units squares can fit on $\mathbb R^2$?

I have thought about this problem for a while and I've come to a solution, however, I am not sure whether my reasoning is good.
Filling the plane with unit squares can be done following this algorithm:
1. Start at $(0.0)$.
2. Place one square to the right, one to the left, one to the right, one to the left and so one, until $\mathbb R \times [0,1]$ has been filled. To fill the mentioned area, we need to repeat this jumping to the left and to the right countably many times for there are as many squares as there are integers.
3. Return to the origin and move one square down - repeat $2$.
4. Repeat $3$ until $\mathbb R \times \mathbb R$ has been filled.

To recap, we have moved from left to right countably many times and up and down countably many times. Therefore, we 'only' need countably many squares to fill the plane.

What do you think of my solution?

Answer 1

We obtain a lattice $\mathbb{Z}^2$ inside $\mathbb{R}^2$ from the covering of unit squares, by just considering the corners. Since $\mathbb{Z}^2$ is countable, also the number of tesselating unit squares is countable.

Answer 2

I think there is a problem when you say "until" you fill $\mathbb{R}\times[0,1]$ since that would take infinite steps (so maybe you would have to use a transfinite induction argument). To fix this you can tile the plane in a "spiral" way.

On the other hand maybe you need to make your question more specific: Can the squares overlap? From your question I understand that the squares can only overlap at their boundary. If that is the case notice that each square will contain an element of the form $(p,q) \in \mathbb{Q}\times \mathbb{Q}$ so the squares are less than all the rationals but more than finite therefore they are countably many. Of course if you allow any kind of overlaping you can fit uncountably many.

It is a non trivial fact that you cannot tile the plane with non-overlaping squares with boundary.