Let $S \subseteq \mathbb{R}^3$ be a smooth surface with orientation given by the unit normal field $N: S \rightarrow \mathbb{S}^2$. $N$ is itself smooth, in the sense of smoothness on $S$ as a $2-$manifold.
Why are the principal curvature smooth (in the sense of manifolds)?
Principal curvatures at $p \in S$ are the eigenvalues of the operator $L_p: T_pS \rightarrow T_pS$, $L_p(v)=-dN_p(v)$, where $dN_p$ is the differential of $N$ at $p$ (in the sense of manifolds).
I know intuitively how the proof should go (?): $L_p$ depends smoothly on $p$ and so do its eigenvalues, hence the principal curvature. The last bit is an extension of "roots of a polynomials depends continuously on its coefficients", fact I'm not sure can be extended to "depends smoothly", and I don't know how to apply here, since I can't translate the dependence of $L_p$ from $p$ into some kind of $2\times 2$ matrix.
I don't know how to proceed since I'm not sure on how to deal with the functions $k_i$. Can you point me to a reference of a proof of this fact, or give me some hint on how to proceed?
$f: S \rightarrow \mathbb{R}$ is smooth in the sense of manifolds if for every parametrization $\phi$ of $S$ we have that $f\circ\phi$ is smooth in the classical way.
$f: S \rightarrow \mathbb{R}^n$ is smooth in the sense of manifolds if every component $f_i$ is smooth in the sense of the above definition.
Given a local parametrization of your surface you can express the principal curvatures as a smooth function of the Gaussian and mean curvature, which depend smoothly on the coefficient of the metric and form coefficients.
Namely, if $\phi$ is the parametrization, the matrix $A\in M_{2,2}(\mathbb{R})$ which represent the endomorphism $dN$ in the coordinate base given by $\phi$ is $$ A= - \begin{vmatrix} E & F \\ F & G \\ \end{vmatrix}^{-1} \begin{vmatrix} e & f \\ f & g \\ \end{vmatrix} $$ where $E, F, G$ and $e, f, g$ are the coefficients of the first and second fundamental form respectively, which depend smoothly on $p$. Now if you take determinant and minus the mean of the trace you get the two curvatures, and finally $$k_{1,2}=H \pm \sqrt{H^2-K}$$ where $K$ is the Gaussian Curvature and $H$ is the mean curvature. Last thing you need to do is to extend this to the whole surface which should be easy "gluing" everything together using the local parametrization
Edit: As Anthony has pointed out in the comments, this expression, $\textit{a priori}$, is smooth only if you stay away from umbilic points. It is not clear to me actually what happens there, see his comment for further details.