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.
If two matrices $\;A,B\;$ are simultaneously diagonalizable (and thus they commute, for example), we get
$$\begin{cases}A=P^{-1}D_1P\\{}\\B=P^{-1}D_2P\end{cases}\;\;\;\implies AB=P^{-1}D_1D_2P$$
and clearly $\;D_1D_2\;$ is also diagonal.
Best Answer
For example, try $A = \pmatrix{1 & 1\cr 0 & 1\cr}$ and $B = \pmatrix{1 & t\cr 0 & 1\cr}$ where $t \ne 0$. $A$ and $B$ both are non-diagonalizable with eigenvalue $1$ and eigenvector $\pmatrix{1\cr 0\cr}$. $AB = \pmatrix{1 & 1+t\cr 0 & 1\cr}$, which is also non-diagonalizable except when $t = -1$, in which case it is diagonalizable.
So you can't always tell whether $AB$ will be diagonalizable by just looking at the diagonalizability, eigenvalues and eigenvectors of $A$ and $B$ separately.