The proof of this property is not so easy as those of the basic properties of eigenvalues and eigenvectors. It can be shown by induction, or by explicit construction (see eg here)
I like to visualize the property in this way:
We know that an hermitian matrix with $n$ distict eigenvalues has $n$ eigenvectors that are not only LI (as in general matrices) but, more than that, orthogonal. We also know that this matrix is diagonalizable, with unitary $U$ (both properties are easy to prove).
Now, if our hermitian matrix happens to have repeated (degenerate) eigenvalues, we can regard it as a perturbation of some another hermitian matrix with distinct eigenvalues.
By a continuity argument, we should see that the matrix perturbation than transforms different (but perhaps close) eigenvalues into coincident ones, cannot make the orthogonal eigenvectors linearly dependent.
Put in other way: an hermitian matrix $A$ with repeated eigenvalues can be expressed as the limit of a sequence of hermitian matrices with distinct eigenvalues. Because all members of the sequence have $n$ orthogonal eigenvectors, by a continuity argument, they cannot end in LD eigenvectors.
This approach leads to a nice intuition, IMO, and it can be formalized. But for a formal proof the other methods are to be preferred.
The multiplicity of an eigenvalue $\lambda$ as a root of the characteristic polynomial of $A$ is called the algebraic multiplicity of $\lambda$. A more geometric characterization of the algebraic multiplicity is the number of time $\lambda$ appears on the diagonal of some upper triangular matrix $B$ which is similar to $A$. Since the characteristic polynomial of $B$ is $p_B(x) = \prod_{i=1}^n (x- b_{ii})$ (because $xI - B$ is upper triangular) and $p_A(x) = p_B(x)$ (because $A$ and $B$ are similar) we see that the multiplicity of $\lambda$ as a root of $p_A(x)$ is the same as the number of times $\lambda$ appears on the diagonal of $B$ (or any other upper triangular matrix which is similar to $A$).
One can use the Jordan form of $A$ which requires proving that any matrix is similar to a matrix in Jordan form but one can also prove directly the much easier result that any matrix with entries in a algebraically closed field is similar to some upper triangular matrix.
Finally, when we are working over an algebraically closed field, the fact that the determinant of $A$ is the product of the eigenvalues (with each eigenvalue counted according to its algebraic multiplicity) follows immediately from the definitions. Since we can factor $p_A(x)$ into linear factors, we can write $p_A(x) = \prod_{i=1}^m (x - \lambda_i)^{a_i}$ and then
$$p_A(0) = \det(xI - A)|_{x = 0} = \det(-A) = (-1)^n \det(A) = (-1)^n \prod_{i=1}^m \lambda_i^{a_i}. $$
Best Answer
Silly example: the zero matrix is clearly normal, but has $$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ as a square root.