Let $A(t)$ be a symmetrical matrix such that each element $A_{ij}(t)$ is a continuous function of a vector $t\in T\subset \mathbb{R}^m$. Are the eigenvalues of $A(t)$ continuous functions of $t$?
I found a few answers that show that this is the case when $T \subset \mathbb{R}$ (for instance this one and this one). Can these results be generalized to $T\subset \mathbb{R}^m$ for $m>1$?
Here is my thought on your problem. Basically, your matrix which is a square and symmetric has elements which depend on the multi-variable vector $\mathbf{t}\in\mathbb{R}^m$. The characteristic polynomial this will generate in $\lambda$ will have coefficients which depend on your vector $\mathbf{t}$. Since finding the eigenvalues comes down to finding the roots of your polynomial there that, to me, would imply that the eigenvalues will be continuous function of $\mathbf{t}$ too. On a simple case thinking of a $2\times 2$ matrix with elements $A_{ij}(\mathbf{t})$, the polynomial would have the form
$$ \lambda^2 -Tr(\mathbf{M}(\mathbf{t}))\lambda + \det\mathbf{M}(\mathbf{t})$$
I think shows clearly that the Eigenvalues would, of course depend on $\mathbf{t}$ since the coefficient clearly depend on $\mathbf{t}$.