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 $\mu(m^{-1}(\lambda)) > 0$.
If $A: H \to H$ is self-adjoint, then $A$ is bounded (Hellinger Toeplitz !).
If $r(A)= \max\{|\lambda|: \lambda \in \sigma(A)\}$ denotes the spectral radius of $A$, then it is well-known that $r(A)=||A||$, if $A$ is self-adjoint.
Hence, if $\sigma(A)=\{0\}$, then $r(A)=0$, hence $A=0.$
Best Answer
Yes, you can calculate the spectral projection for each eigenvalue $\lambda$ by integrating the resolvent in a small contour around $\lambda$ that avoids all of the other eigenvalues $$P_\lambda = \frac{1}{2\pi i} \int_C (T-z I)^{-1} \, dz.$$ The Hilbert space will then be the direct sum of the spectral subspaces corresponding to the spectral projections. Isolated elements of the spectrum are always eigenvalues.