If the spectrum is bounded, then by means of holomorphic functional calculus we can extract a projection:
$\displaystyle P = \frac{1}{2\pi i} \intop_\gamma (\lambda - T)^{-1} d\lambda$,
where $\gamma$ encloses the spectrum. Intuitively, this should be thought of as separating the "bounded part" $TP$ (which is indeed bounded, since $T\left(\lambda-T\right)^{-1}=\lambda\left(\lambda-T\right)^{-1}-1$) and the part $T-TP$, which has empty spectrum on $\mathbb{C}$ when restricted to $\ker P$, but should be thought of as having a point at infinity, since indeed its inverse is a bounded operator with spectrum $\{0\}$.
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
Assume $T$ has empty spectrum. Then $T$ is invertible, $T^{-1}$ is a bounded selfadjoint operator and, for $\lambda \ne 0$, $$ (T^{-1}-\lambda I) =(I-\lambda T)T^{-1}=\lambda(\frac{1}{\lambda}I-T)T^{-1} $$ has bounded inverse $$ \frac{1}{\lambda}T\left(\frac{1}{\lambda}I-T\right)^{-1} $$ So $\sigma(T^{-1})=\{0\}$ because only $\lambda=0$ can be in the spectrum, and it cannot be empty. But that implies $T^{-1}=0$, which is a contradiction.