Suppose $T$, $S$ are bounded operators on $l_2$, $a_n\to 0$ a sequence of complex numbers with the property that for any $n\in\mathbb{N}$, $T+a_nS$ has discrete spectrum and non-empty point spectrum. Does it follow that $T$, as the norm limit of $T+a_nS$ also has non-empty point spectrum?
If the answer would be "no" for any $S$, would it change if we take $S$ compact or finite rank?
An approach to this question, but not an answer, and too long for a comment.
Say, for simplicity, that $T$ is quasinilpotent. Then $r:\mathcal{L(l_2)}\to:[0,\infty)$ is continuous at $T$. This means that $r(T+a_nS)\to 0$. Pick $\lambda_n$ to be eigenvectors with norm $1$ eigenvalues $x_n$, for each of $T+a_nS$. Then $a_n\to0$ as well as $\lambda_n\to 0$. Since $(x_n)$ bounded, WLOG assume $x_n\stackrel{w}{\to}x$ (we are in $l_2$). Taking $w$-limits in the relation
$$ Tx_n+a_nSx_n=\lambda_nx_n $$
it follows that $Tx=0$. If we could ensure that $x\neq 0$, this would mean that $0$ is an eigenvalue for $T$, and the question would have an affirmative answer. I don't know how to make $x\neq 0$, or even if it is possible. I don't think that $S$ compact makes any difference.