Does the limit of this sequence of operators have infinitely many eigenvalues?

Suppose that I have a sequence of compact, injective operators $$\{T_\delta\}_{\delta>0}$$ on a Hilbert space $$H$$ such that each operator $$T_\delta$$ has infinitely many eigenvalues. My question is the following.

Question: If $$T$$ is a compact, injective operator on $$H$$ and $$T_\delta\to T$$ in operator norm as $$\delta\to0^+$$, then must $$T$$ have infinitely many eigenvalues?

I know that each eigenvalue of $$T$$ must be a limit of eigenvalues of $$T_\delta$$, but I am not sure how to establish the converse in the above sense. According to Convergence of spectrum with multiplicity under norm convergence, any $$\lambda\in\mathbb{C}$$ which is the limit point of eigenvalues of $$T_\delta$$ (with multiplicity taken into account) must be an eigenvalue of $$T$$, but I am not sure how to guarantee that infinitely many such points exist. I have mainly used Kato's book Perturbation Theory for Linear Operators as a reference, but I have been unable to find the right result.

An example of a compact injective operator on $$\ell^2(\mathbb Z)$$ (i.e. square-summable two-sided sequences) with no eigenvalues is $$T$$ defined by $$(Tx)_i = x_{i+1}/(|i|+1)$$ This is the norm limit of compact injective operators $$T_N$$ where $$(T_N x)_i = \cases{x_{i+1}/(|i|+1) & if |i| \le N\cr x_{-N}/N & if i = N+1\cr x_{i}/(|i|+1) & otherwise\cr}$$ which have infinitely many eigenvalues.