Given the edit, the assertion is true in a more general setting (namely, you don't need to fix the degeneracies) and it is contained in theorem 1.10 of Kato Perturbation theory for linear operators:

Theorem (Kato) Let $T(x)$ be an $n\times n$ matrix depending analytically on $x$ in a domain $D_0$ of the complex plane. Let $x_ \in D_0$ (note that $x_0$ may be an exceptional point, i.e., there may be degenerate eigenvalues) and let there exist a sequence $x_n$ converging to $x_0$ such that $T(x_n)$ is normal for $n=1,2,\ldots$. Then all the eigenvalues $\lambda_j$ (and eigenprojectors $P_j$) are holomorphic at $x=x_0$ and there are no nilpotent terms in the spectral decomposition.

The proof essentially uses the fact that, if there was a singularity (which must be of the form $x^{(1/p)}+\ldots$ for some $p$), then the norm of the eigenprojectors would be unbounded, but for normal operators that's impossible.

Edit I have erroneously interpreted the edit. The following only addresses the smoothness of the mapping $x \mapsto (\mathrm{eigenvalues})$.

Given the edit, the assertion is true in a more general setting (namely, you don't need to fix the degeneracies) and it is contained in theorem 1.10 of Kato Perturbation theory for linear operators :

Theorem (Kato) Let $T(x)$ be an $n\times n$ matrix depending analytically on $x$ in a domain $D_0$ of the complex plane. Let $x_ \in D_0$ (note that $x_0$ may be an exceptional point, i.e., there may be degenerate eigenvalues) and let there exist a sequence $x_n$ converging to $x_0$ such that $T(x_n)$ is normal for $n=1,2,\ldots$. Then all the eigenvalues $\lambda_j$ (and eigenprojectors $P_j$) are holomorphic at $x=x_0$ and there are no nilpotent terms in the spectral decomposition.

The proof essentially uses the fact that, if there was a singularity (which must be of the form $x^{(1/p)}+\ldots$ for some $p$), then the norm of the eigenprojectors would be unbounded, but for normal operators that's impossible.