This is a classic double counting problem. Name the colours $C_1, C_2, \ldots, C_{100}$. We will say that a row or column is $j$-good if it has at least one square of colour $C_j$ on it. We have to prove that there is a row or column which is $j$-good for at least $10$ different values of $j$. For a fixed $j$, suppose there are $a$ rows that are $j$-good and $b$ columns that are $j$-good. Since all $C_j$ squares on the board are on the intersection of a $j$-good row and a $j$-good column, it follows that the number of these squares is at most $ab$. Hence $ab \geq 100$, which implies (say, by AM-GM) $a+b \geq 20$. Now consider all pairs $(j,\ell)$ where $1 \leq j \leq 100$ and $\ell$ is a row or column that is $j$-good. By the previous remarks, there are at least $100 \cdot 20 = 2000$ of such pairs. Since there are $2 \cdot 100 = 200$ possible rows and columns $\ell$, it follows that at least one of them is paired with at least $\frac{2000}{200} = 10$ colours. $\blacksquare$