Suppose I have a $2 \times 2$ Hermitian matrix-valued function $m$ defined on $\mathbb{R}^{2}$ with entries $m_{jk} \in C^{\infty}(\mathbb{R}^{2})$. Denote by $m_{+}$ and $m_{-}$ the greater and smaller eigenvalues, respectively. Is it true that $m_{\pm}$ will also be $C^{\infty}$ functions?
If not, can you give an example for which this does not hold?
$m_\pm$ need not be smooth. Here's an example with a matrix depending on one real parameter. Put $$M = M(t) = \begin{pmatrix}1&t\\t&1\end{pmatrix}\quad for \quad t \in \mathbb R. $$ $M$ is hermitian and depends smoothly on $t$. For any given $t$, we have the eigenvector $\begin{pmatrix}1\\1\end{pmatrix}$ with eigenvalue $\alpha = \alpha(t) = 1 + t$ and the eigenvector $\begin{pmatrix}1\\-1\end{pmatrix}$ with eigenvalue $\beta = \beta(t) = 1 - t$. Note that for $t \neq 0$, $M(t)$ has two different eigenvalues, while for $t = 0$, $M(0)$ has the eigenvalue $0$ with multiplicity $2$ since the eigenvectors are linearly independent. Now, $$ m_+(t) = \max\{\alpha(t),\beta(t)\} = \max\{1+t,1-t\} = 1+\max\{t,-t\} = 1+|t| $$ and similarly $$ m_-(t) = \min\{\alpha(t),\beta(t)\} = \min\{1+t,1-t\} = 1+\min\{t,-t\} = 1-|t| $$ This shows that $m_+(t)$ and $m_-(t)$ do not depend smoothly on $t$, although there is a smooth parametrization of the eigenvalues, namely $\alpha(t)$ and $\beta(t)$.
If you insist on a counterexample depending on two real parameters, you can put $$ \widetilde M = \widetilde M(t,u) = M(t) \quad for \quad t,u \in \mathbb R. $$