Suppose $V$ is a infinite-dimensional separable real Hilbert space and $$ T\,\colon\,I \to B(V), $$ where $I = [0,a) \subset \mathbb{R}$ and $B(V)$ denotes the space of bounded linear operators on $V$. The short-hand notation is $T(\epsilon) \equiv T_{\epsilon}$ for any $\epsilon \in I$.
Suppose further that $T$ is continuous (possibly Frechet differentiable?) and that $T_0$ is simply given by $$T_0v = cv,\quad(1)$$ where $c > 0$. Final assumptions are that for any $\epsilon \in I$, $T_{\epsilon}$ is self-adjoint and is an isomorphism and for $\epsilon > 0$ it is not of the form given in (1).
(Side note: I am looking at a specific example but I wanted to look at the problem more abstractly, perhaps some of the 'assumptions' either follow from the others or are not needed).
I would like to be able to characterise the set of eigenvalues and eigenvectors of $T_{\epsilon}$ and how they depend on the parameter $\epsilon$. Trivially, the form of $T_0$ implies that $c$ is the sole eigenvalue of $T_0$ and its corresponding eigenspace is the whole of $V$.
It seems fairly natural that the eigenvalues will be continuous functions of $\epsilon$ since $T$ is a continuous function and so if $\lambda_\epsilon$ is an eigenvalue of $T_\epsilon$ with the corresponding eigenvector $v_{\epsilon}$ then $$ (\lambda - c)v_{\epsilon} = (T_{\epsilon}-T_0)v_{\epsilon} \implies |\lambda - c|\|v_{\epsilon}\| \leq \|(T_{\epsilon}-T_0)\|\|v_{\epsilon}\| \implies |\lambda - c| \leq C\epsilon, $$ with the first equality following from the fact that $v_{\epsilon}$ is also an eigenvector for $T_0$. But what gives us the certainty that $T_{\epsilon}$ has any eigenvalues at all? How many eigenvalues can it have? Since it is self-adjoint and $V$ is separable, there can be at most countably many eigenvalues, so it is tempting to conjecture that the set of eigenvalues is $$(\lambda_1(\epsilon),\lambda_2(\epsilon),\dots)$$ with each entry being a continous function of $\epsilon$ with $\lambda_i(0) = c$ and possibly $\lambda_i(\epsilon) = \lambda_j(\epsilon)$ for some $i\neq j$. A further conjecture would be that each $\lambda_i(\epsilon)$ can be associated with one eigenvector $v_i$ such that $(v_i,v_j) = 0$ and the closure of ${\rm span}(v_1,v_2,\dots)$ is the whole of $V$. For that it feels like we should employ Spectral Theorem, so in particular perhaps establish that $ (T_{\epsilon} - T_0)$ is a compact operator? Is it? I do not quite see it.
I will humbly appreciate any potential comments, I am quite confused here, in particular the notion of orthonormal basis for infinite-dimensional Hilbert space seems a bit elusive. As I mentioned in the side-note, I am keen to consider this more broadly, so perhaps if one of the assumptions makes the problem trivial (perhaps the fact that $T_{\epsilon}$ is an isomorphism?), please feel free to comment more broadly.
If $V$ is finite dimensional, then $T(\varepsilon)$ has real eigenvalues for all $\varepsilon$ and these indeed will depend continuously on the parameter, assuming that $T$ depends smoothly on it. However, you can't expect more than Lipschitz continuous dependence, due to the possibility multiple eigenvalues. Find out more under the term "matrix perturbation theory".
If $V$ is infinite dimensional, then the fundamental problem is that $T(\varepsilon)$ need not have eigenvalues for $\varepsilon > 0$. Your question is not well posed in this case.
An example is given by the operator $T(\varepsilon)$ acting on $V = L^2(\mathbb{R})$, defined by $$ \left(T(\varepsilon)f\right)^\tilde {}(s) = \tilde{f}(s) \cdot \frac{2 + \varepsilon |s|}{1 + \varepsilon |s|} $$ where $\tilde g$ is the Fourier transform of $g$. Find out more under the general term "spectral theory of operators".