= Solution
For the subsequent <essential spectrum> arguments, the discrete-spectrum definition must use $\operatorname{ran}(L-\lambda I)$ closed, rather than the printed unshifted range. With that correction, $0\notin\Sigma_{\mathrm e}(L)$ means exactly that $N=\ker L$ is finite-dimensional and $\operatorname{ran}L$ is closed; this includes the case $0$ is in the <resolvent set>.
A <closed-range bound on the kernel complement> supplies the useful equivalence
$$
\operatorname{ran}L\text{ closed}\iff\exists b>0:\ \|Lv\|\ge b\|v\|\quad(v\in N^\perp).
$$
For the forward implication, $L:N^\perp\to\operatorname{ran}L$ is a bounded bijection between <Banach spaces>, so the <bounded inverse theorem> applies. For the reverse implication, any <Cauchy sequence> of image points has a <Cauchy sequence> of preimages in $N^\perp$, and completeness gives a preimage of its limit.
Now decompose a bounded sequence as $f_n=p_n+v_n$ with $p_n\in N$, $v_n\in N^\perp$. If $Lf_n$ converges, the lower bound makes $(v_n)$ a <Cauchy sequence>. The <finite-dimensional vector space> $N$ makes the bounded $p_n$ have a convergent subsequence. Their sum has a norm-convergent subsequence.
Conversely, if every bounded sequence with convergent images has a norm-convergent subsequence, the kernel cannot be infinite-dimensional: an <orthonormal sequence> in it would have zero images and no convergent subsequence. If the range were not closed, the lower-bound equivalence would provide unit vectors $v_n\in N^\perp$ with $Lv_n\to0$. Any norm limit would lie in both $N$ and $N^\perp$, hence be zero, contradicting its unit norm. This proves \b[the required <sequential properness for a self-adjoint operator> equivalence].
The spectral-shift repair is essential for later parts. On $\ell^2$, take $L=\operatorname{diag}(2,1,1/2,1/3,\ldots)$. Its range is not closed, but $2$ is an isolated eigenvalue with one-dimensional eigenspace and closed shifted range. The printed definition would incorrectly place $2$ in the <essential spectrum>, although no <singular Weyl sequence> exists there: on the complement of that eigenspace, $\|(L-2I)v\|\ge\|v\|$.
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