Does an unbounded operator $T$ with non-empty spectrum have an unbounded spectrum?

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 $\{0\}$.

Weidmann gives a proposition to check for boundedness by investigating the numerical range (see proposition 2.51 in german version of Weidmann's 'Lineare Operatoren in Hilberträumen'). This makes sense since the numerical range is the natural object when studying boundedness rather than the spectrum...

It states that for either (a) arbitrary operators on complex Hilbert spaces or (b) not necessarily densely defined symmetric operator operators on arbitrary Hilbert spaces we have:

-->The operator is bounded iff its numerical range is bounded!

For not necessarily bounded normal densely defined operators we have:
(Relaxing to closed ones fails in general as seen in numerical range vs spectrum.)

-->The spectrum is contained in the closure of the numerical range!

So the scheme is as follows: $$A\text{ bounded}\iff\mathcal{W}(A)\text{ bounded}$$ $$N^*N=NN^*:\quad\sigma(N)\subseteq\overline{\mathcal{W}(N)}$$

But apart from that there's (unfortunately) nothing much to say I guess.

By the way if you're more concerned about little subtlties like denseness etc. Weidmann is a good book for. He tries to keep these aspects always (annoyingly or luckily) in discussion.