We can do this by (degenerate) first order perturbation theory. Let's take $D=1+\epsilon C$, with $\epsilon\not=0$ small and $C$ also diagonal. Then $$ DAD = A +\epsilon(CA+AC) + O(\epsilon^2) . $$ Let's focus on a multiple eigenvalue $\lambda$, with eigenvector basis $v_1,\ldots ,v_k$. Perturbation theory says that in first order, the $k$ copies of $\lambda$ will be moved to $\lambda+\epsilon\mu_j$. Here, the $\mu_j$ are the eigenvalues of $CA+AC$, compressed to $L(v_1,\ldots, v_k)$. In other words, they are the eigenvalues of the $k\times k$ matrix with entries $$ \langle v_j, (CA+AC) v_m\rangle = 2\lambda \langle v_j, Cv_m\rangle . $$

Now we only need to make sure that not all the $\mu_j$ are equal to one another to at least remove some of the degeneracies. This we can of course easily do, for example by taking $C$ as a suitable rank $1$ matrix.

We have reduced $\sum (n(\lambda)-1)$ (with $n(\lambda)$ denoting the multiplicity of $\lambda$) by at least $1$; notice here that for any fixed $C$, our perturbation will never introduce new degeneracies, provided we take $\epsilon$ small enough. Finally, repeating this whole step will eventually get us to a matrix with no degeneracies.

Note also that this argument does not work for $\lambda=0$, which is exactly as it must be since of course a multiple eigenvalue $\lambda=0$ can not be moved by such a multiplicative perturbation.

We can do this by (degenerate) first order perturbation theory. Let's take $D=1+\epsilon C$, with $\epsilon\not=0$ small and $C$ also diagonal. Then $$ DAD = A +\epsilon(CA+AC) + O(\epsilon^2) . $$ Let's focus on a multiple eigenvalue $\lambda$, with eigenvector basis $v_1,\ldots ,v_k$. Perturbation theory says that in first order, the $k$ copies of $\lambda$ will be moved to $\lambda+\epsilon\mu_j$. Here, the $\mu_j$ are the eigenvalues of $CA+AC$, compressed to $L(v_1,\ldots, v_k)$. In other words, they are the eigenvalues of the $k\times k$ matrix with entries $$ \langle v_j, (CA+AC) v_m\rangle = 2\lambda \langle v_j, Cv_m\rangle . $$

Now we only need to make sure that not all the $\mu_j$ are equal to one another to at least remove some of the degeneracies. This we can of course easily do, for example by taking $C$ as a suitable rank $1$ matrix.

We have reduced $\sum (n(\lambda)-1)$ (with $n(\lambda)$ denoting the multiplicity of $\lambda$) by at least $1$; notice here that for any fixed $C$, our perturbation will never introduce new degeneracies, provided we take $\epsilon$ small enough. Finally, repeating this whole step will eventually get us to a matrix with no degeneracies other than possibly $\lambda=0$.

Note also that this argument does not work for $\lambda=0$, which is exactly as it must be since of course a multiple eigenvalue $\lambda=0$ can not be moved by such a multiplicative perturbation.