 Solution

The spectrum of a bounded operator consists of $\lambda$ for which $T-\lambda I$ has no bounded two-sided inverse. The point spectrum consists of eigenvalues, so $\ker (T-\lambda I)\ne 0$. The approximate point spectrum consists of $\lambda$ admitting unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$; equivalently $T-\lambda I$ is not bounded below.

If $T-\lambda I$ is invertible, its inverse bounds it below and its Banach-space adjoint is invertible. Thus neither alternative on the right of the claimed union occurs. Conversely suppose $\lambda$ belongs to neither term and write $S=T-\lambda I$. There is $c>0$ with $\Vert Sx\Vert \ge c\Vert x\Vert$, so $S$ is injective and has closed range: convergence of $S{x}_n$ makes $x_n$ Cauchy, and completeness supplies a preimage of its limit. Also $\ker S^{\prime }=0$. The allowed annihilator identity makes the range dense, hence equal to $X$. Its inverse has norm at most $1/c$. This proves the spectrum decomposition through the Banach-space adjoint:For the right shift on ${\ell }^1$, $\Vert T\Vert =1$, so the allowed norm bound puts its spectrum in the closed unit disc. Under the linear pairing with ${\ell }^{\infty }$, its Banach-space adjoint is the left shift $T^{\prime }(b_1,b_2,\dots )=(b_2,b_3,\dots )$. For each $|\lambda |\le 1$ the bounded nonzero sequence $(1,\lambda ,{\lambda }^2,\dots )$ is an eigenvector of $T^{\prime }$ with eigenvalue $\lambda$. The union identity gives the reverse inclusion. Hence $\sigma (T)=\{\lambda :|\lambda |\le 1\}$. This Banach-space adjoint uses the complex-linear dual, so its eigenvalue is $\lambda$ itself.