Eigenvalues are only Hölder continuous

The matrix. $M_\varepsilon=J+\varepsilon E_{n1}$ has $(M_\varepsilon)_{i,i+1}=1$ for $i<n$ and $(M_\varepsilon)_{n,1}=\varepsilon$, all other entries $0$. Equivalently $M_\varepsilon e_1=\varepsilon e_n$ and $M_\varepsilon e_{j}=e_{j-1}$ for $j\ge2$.

Characteristic polynomial. Let $x=(x_1,\dots,x_n)$ be an eigenvector for $\mu$: $M_\varepsilon x=\mu x$ reads $x_{i+1}=\mu x_i$ for $i<n$ and $\varepsilon x_1=\mu x_n$. Hence $x_k=\mu^{k-1}x_1$, $x_1\ne0$, and $\varepsilon x_1=\mu\cdot\mu^{n-1}x_1$, so $\mu^n=\varepsilon$. The $n$ numbers $\mu_k=\varepsilon^{1/n}e^{2\pi ik/n}$ are distinct, so
$$\chi_{M_\varepsilon}(t)=t^n-\varepsilon.$$

Conclusion. All eigenvalues of $M_\varepsilon$ have modulus $\varepsilon^{1/n}$, while those of $M_0=J$ are $0$. The perturbation has operator norm $\|M_\varepsilon-M_0\|=\varepsilon$, so
$$\max_k|\mu_k-0|=\varepsilon^{1/n}=\|M_\varepsilon-M_0\|^{1/n}\gg\|M_\varepsilon-M_0\|\qquad(\varepsilon\to0,\ n\ge2).$$
No inequality $|\mu-\mu'|\le C\|M-M'\|$ can hold near $J$.

Example. $n=2$, $\varepsilon=10^{-8}$: $\begin{pmatrix}0&1\\10^{-8}&0\end{pmatrix}$ has eigenvalues $\pm10^{-4}$, four orders of magnitude larger than the perturbation.

Remark. The exponent $1/n$ is optimal in general (Bauer-Fike type bounds for non-diagonalisable matrices give Hölder $1/n$). For normal matrices the dependence is Lipschitz with constant $1$, as shown in the Bauer-Fike problem below.