The Bauer-Fike theorem

Trivial case. If $\mu\in\{d_i\}$ the left side is $0$ and there is nothing to prove. Assume $\mu\notin\{d_i\}$, so $D-\mu I$ is invertible.

Reformulation. Let $x\ne0$ with $(A+E)x=\mu x$. Then $(A-\mu I)x=-Ex$. Substituting $A=VDV^{-1}$ and $y=V^{-1}x\ne0$:
$$V(D-\mu I)V^{-1}x=-Ex\ \Longrightarrow\ (D-\mu I)y=-V^{-1}EVy.$$

Inverting $D-\mu$.
$$y=-(D-\mu I)^{-1}V^{-1}EV\,y.$$
Take norms and use submultiplicativity:
$$|y|\le\|(D-\mu I)^{-1}\|\,\|V^{-1}\|\,\|E\|\,\|V\|\,|y|.$$

Conclusion. Dividing by $|y|>0$ gives $1\le\|(D-\mu I)^{-1}\|\,\kappa(V)\,\|E\|$. For a diagonal matrix, $\|(D-\mu I)^{-1}\|=\max_i\frac1{|d_i-\mu|}=\frac1{\min_i|d_i-\mu|}$. Therefore
$$\min_i|d_i-\mu|\le\kappa(V)\|E\|.$$

Example. $A=\begin{pmatrix}1&1\\0&1+\delta\end{pmatrix}$ has eigenvectors with $\kappa(V)\approx2/\delta$ (nearly parallel eigenvectors), so a perturbation of size $\eta$ can move the eigenvalues by about $\sqrt{\eta}$ when $\delta\approx\sqrt\eta$: the bound is quantitatively sharp as $A$ approaches a Jordan block.

Remark. The theorem quantifies how ill-conditioned the eigenvalues of a nearly non-diagonalisable matrix are, and recovers the Hölder phenomenon of the Jordan block ($\kappa\to\infty$).