Let the sides of the squares be equal to $a_i$ for $i = 1, 2, \ldots, N$ ($N$ is the number of squares).

If some $a_k > 1$ then the $k$th square will cover the unit square. Now let's assume the case when $a_i < 1$ for all $i$.

Every number $a_i$ must satisfy $2^{-k_i} \le a_i < 2^{-k_i+1}$ for some integer $k_i$. Now let's decrease every $i$th square to the square with side $2^{-k_i}$. Then its area would decrease by at most $4$ times. Therefore the area of all squares will be greater than $1$.

Now let's prove we can tile the unit square fully with the new squares. Let's divide the unit square into $4$ squares of side $\frac{1}{2}$. First place the squares with side $\frac{1}{2}$, if they exist. Then on the non-tiled squares with side $\frac{1}{2}$, if they exist, place the squares with side $\frac{1}{4}$, if they exist, dividing each non-tiled square with side $\frac{1}{2}$ into $4$ equal squares. We will continue this procedure for $k = 3, 4, \ldots$ by placing squares with side $\frac{1}{2^k}$ on the non-tiled squares with side $\frac{1}{2^{k-1}}$, in turn dividing them into $4$ equal squares.

Finally, because the sum of areas of squares is greater than $1$, then on some step, we will cover the square. Then increasing the $i$th square to the square with side $a_i$, we will get the tiling of the unit square with the given squares.