Is the following conclusion true? Suppose $A,B$ are $n\times n$ complex Hermitian matrices. Then there exists real analytic functions $\lambda_i:\mathbb R\to \mathbb R$ where $1\leq i\leq n $ such that for each $t\in\mathbb R$, the set $\{\lambda_1(t)\dots,\lambda_n(t)\}$ is the set of all eigenvalues of $A+tB$ counting multiplicities and with respect to some orthonormal basis $A+tB=\text{diag}(\lambda_1(t),\dots,\lambda_n(t)).$
If this is true, is it also true for $A,B$ are Hermitian compact operators on $\ell_2$?
$t\in \mathbb{R}\mapsto A+tB$ is analytic with values in the set of hermitian matrices. Then your conjecture is true.
cf. for the real symmetric case, my post in
Do eigenvalues depend smoothly on the matrix elements of a diagonalizable matrix?
More generally, for the hermitian case or the operators, cf.
https://arxiv.org/pdf/1111.4475.pdf
EDIT. Answer to the OP. The considered theorem is valid fo one real parameter. There are counter-examples when we are dealing with $2$ parameters.
EDIT. Answer to the OP.
Note also that when $A,B$ are invertible and $n$ is odd, "there is always some real $t$ s.t. $A+tB$ is not invertible". This last point is useless -I don't know why I wrote this-; there is a global analytic parameterization over $(-\infty,+\infty)$:
$A+tB=U(t)^*diag((\lambda_i(t))_i)U(t)$.