We number rows numbered from $1$ to $8$ (from top to bottom), and columns numbered from $1$ to $8$ (from left to right). Suppose by contradiction that each $2 \times 2$ square board does not contain exactly one colored square.

Obviously there must also be an empty column, let's say it is the $i$ column with $1 \le i \le 8$. Let's consider the left column (if any) of this $i$ column, that is, the $(i-1)$-th column. If on that $(i-1)$-th column, there is a colored square. Let's consider the corresponding $2 \times 2$ square containing $2$ squares on $i$-th column (which are two uncolored squares) and also containing the $\times$ cell. Thus, the remaining cell in $(i-1)$-th column must also be colored because otherwise the $2 \times 2$ table contains exactly $1$ colored cell, which contradicts the assumption.

Continuing like that, consider the squares containing $2$ cells in column $i$ and containing a colored cell in column $i-1$, then by similar reasoning, the remaining cell is also colored. This shows that the entire column $i-1$ must be colored, that is, there are $8$ colored cells, which is absurd. From there, we can deduce that column $i-1$ is an empty column. The same reasoning goes for column $i+1$. From there, we can continue to expand to columns $i-2, i-3, \dots, 1$ and columns $i+2, i+3, \dots, 8$, which are also empty columns. Therefore, the entire table is not colored, which is also absurd. This proves the given assertion. $\square$