Elementary question about Cayley Hamilton theorem and Zariski topology

This map is not only continuous. It is a actually a morphism of algebraic varieties. To see this note that the characteristic polynomial of a matrix is a polynomial whose coefficients are polynomials in the entries of the matrix. This is because it is equal to $\det(tI-A)$, and the determinant of a matrix is a polynomial in its entries. Also, the discriminant of a polynomial is again, a polynomial in its coefficients, so we see that this map is a polynomial in the entries of $A$.

$f$ is a polynomial hence it's continuous wrt. Zariski topology. You are almost done:you need to show that the complement of $f^{-1}(0)$ is non-empty, and this you know as there is matrix with different eigenvalues (eg. a suitable diagonal matrix)