Just to give you some intuition, consider the case of finite-dimensional space.
Let $A$ be a self-adjoint operator on $\mathbb C^n.$ That is $Ae_k=\lambda_k e_k,\ \lambda\in\mathbb R$ for some ONB $e_k,\ k=1,\dots, n.$ Then:
$A$ has simple spectrum iff $\lambda_k\neq \lambda_m$ for all $k,m$ iff there is a cyclic vector $v\in\mathbb C^n$ for $A.$
(Cyclic means that $\{p(A)v\mid p\ \mbox{polynomial}\}$ is dense in $\mathbb C^n$. In this case you can take $v=(1,1,\dots,1)$).
Thus you can define a bounded s.-a. operator to have a simple spectrum if it has a cyclic vector. (You can easily check that multiplication operator $f(x)\mapsto xf(x)$ on $L^2(\mathbb R,\mu)$ has a cyclic vector $v\equiv 1,$ $\mu$ is Borel measure with compact support).
Another characterization: the spectrum of $A$ is simple iff there exists an $x_0\in H$ such that the linear hull of $$\{ (E_{\lambda_1}-E_{\lambda_2})x_0\ : \ \lambda_2<\lambda_1\}$$ is a dense subset of $H$. For the multiplication operator $f(x)\mapsto xf(x)$ it means that the set of step functions is dense in $L^2(\mathbb R,\mu).$
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 $\mu(m^{-1}(\lambda)) > 0$.
Best Answer
Every non-zero spectral value of a self-adjoint compact operator (see for instance Rudin, Functional Analysis, p109) is an eigenvalue, therefore, you have : $$c(T) \subset \{0\}$$
It is possible that $c(T)=\emptyset$ ($T=0$ for instance), but it can also be not empty. Indeed, suppose $\dim(H) = \infty$ and $H$ separable. Pick your favourite orthonormal basis $(e_n)$ and define $T$ by $$T(e_n)=\dfrac{1}{n+1}e_n$$ Then $T$ is compact (as a limit of finite rank operators) and injective, its range contains the orthonormal basis (hence is dence), but not surjective, thus $c(T)=\{0\}$.
If you impose $\ker(T) \neq \{0\}$ and $\dim(H)=\infty$, then $0$ must belong to the spectrum of $T$ but, since $T$ is not injective, $c(T)=\emptyset$.