I do not know a reference, but here is an easy argument. Consider the space $overline{M}$ of pairs $(A,lambda)$ where $A$ is a (self-adjoint) $N times N$ matrix and $lambda$ is a root of the characteristic polynomial of $A$. Over the open subset of $overline{M}$ lying above the operators with distinct eigenvalues, the projection onto the first factor is clearly a proper smooth map, with finite fibers; in other words, the space $overline{M}$ is a covering space over this open set. Since everything in sight is differentiable, the result that you want follows from the lifting of paths in the base. You need to choose an eigenvalue for the lifting to be unique, but you seem to know which one to choose!
Btw, you only need the eigenvalue $lambda$ to be simple for the same argument to work. In this case you can use the Implicit Function Theorem to lift the path, since the assumption on the simplicity of the root imply that the differential at the point is surjective.
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