Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e.
$$(Tf)(x) = x\cdot f(x).$$
Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint.
Clearly $T$ has no eigenvalues, since
$$(T - \lambda I)f = 0 \iff (x-\lambda)\cdot f(x) = 0 \quad \text{a.e.},$$
and since $x-\lambda$ is nonzero for all but at most one $x\in [0,1]$, it follows that $f(x) = 0$ for almost all $x$.
We have $\sigma(T) = [0,1]$, as is easily seen - for $\lambda \in \mathbb{C}\setminus [0,1]$, the function $x \mapsto \frac{1}{x-\lambda}$ is bounded on $[0,1]$.
The reason that $T$ has no eigenvalues is that for all $\lambda\in \mathbb{C}$ the set $\operatorname{id}_{[0,1]}^{-1}(\lambda)$ has measure $0$. If $m \colon [0,1] \to \mathbb{R}$ is a bounded measurable function such that $\mu(m^{-1}(\lambda)) > 0$ for some $\lambda$, then $\lambda$ is an eigenvalue of $T_m \colon f \mapsto m\cdot f$, any function $f\in H$ that vanishes outside $m^{-1}(\lambda)$ is an eigenfunction of $T_m$ then. Conversely, if $T_m$ has an eigenvalue $\lambda$ and $f \neq 0$ is an eigenfunction to that eigenvalue, then, since $(m(x) - \lambda)f(x) = 0$ almost everywhere, it follows that $f$ vanishes almost everywhere outside $m^{-1}(\lambda)$, and since $f\neq 0$, it further follows that $\mu(m^{-1}(\lambda)) > 0$.
In the following for a linear operator $T$ in $\ell^2(\Bbb Z)$, $\rho(T)$ denotes the resolvent set of $T$, i.e., the set of all $\lambda\in\Bbb C$ for which the operator $T-\lambda = T-\lambda I$ is invertible. Also, $\sigma(T)$ denotes the spectrum of $T$, i.e., $\sigma(T) = \Bbb C\setminus\rho(T)$.
For $\lambda\in\rho(L)$ you have $L+R-\lambda = (I + R(L-\lambda)^{-1})(L-\lambda)$. We know that $\|(L-\lambda)^{-1}\| = \tfrac 1{dist(\lambda,\sigma(L))}$. So, for $dist(\lambda,\sigma(L))$ large enough (note that $\sigma(L) = \{n : n\in\Bbb Z\}$) we have $\|R(L-\lambda)^{-1}\| < 1$ and thus $L+R-\lambda$ is invertible for these $\lambda$. So, $\rho(L+R)\neq\emptyset$. Now, for $\lambda\in\rho(L+R)\cap\rho(L)$,
$$
(L+R-\lambda)^{-1} = (L-\lambda)^{-1}[I + R(L-\lambda)^{-1}]^{-1}.
$$
Now, we also (hopefully) know that $(L-\lambda)^{-1}$ is a compact operator. So $(L+R-\lambda)^{-1}$ is compact. Hence, the resolvent of $L+R$ is compact, which means that $\sigma(L+R)$ is discrete.
For part (b) invoke the Fourier transform $\mathcal F : C(\Bbb T)\to\ell^2(\Bbb Z)$.
Best Answer
Ok, found a solution. Let $\lambda$ be a number such that $w-\lambda Id$ is injective but not invertible. So, we want to show that also $v-\lambda Id=w-u-\lambda Id$ is not invertible.
By absurd, if it was invertible, we would be able to write: $$ w-\lambda Id=w-u-\lambda Id +u= (w-u-\lambda Id)(Id+u')$$ with $u':=(w-u-\lambda Id)^{-1}u,$ a compact operator which exists by invertibility of $(w-u-\lambda Id).$
Because $w-\lambda Id$ is injective, so it is $Id+u',$ and by compactness of $u'$ we obtain by a well-known lemma that $Id+u'$ is also surjective.
But this would mean that $w-\lambda Id$ is surjective, and thus invertible, which leads to a contradiction.