= 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 $\ker(T-\lambda I)\ne0$. The <approximate point spectrum> consists of $\lambda$ admitting <unit vectors> $x_n$ with $\|(T-\lambda I)x_n\|\to0$; 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 $\|Sx\|\geq c\|x\|$, so $S$ is injective and has closed range: convergence of $Sx_n$ makes $x_n$ Cauchy, and <completeness> supplies a preimage of its limit. Also $\ker S'=0$. 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>:
$$
\boxed{\sigma(T)=\operatorname{ap}\sigma(T)\cup\operatorname p\sigma(T').}
$$
For the right shift on $\ell^1$, $\|T\|=1$, 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 $T'(b_1,b_2,\ldots)=(b_2,b_3,\ldots)$. For each $|\lambda|\leq1$ the bounded nonzero sequence $(1,\lambda,\lambda^2,\ldots)$ is an <eigenvector> of $T'$ with <eigenvalue> $\lambda$. The union identity gives the reverse inclusion. Hence \b[$\sigma(T)=\{\lambda:|\lambda|\leq1\}$]. This <Banach-space adjoint> uses the complex-linear dual, so its <eigenvalue> is $\lambda$ itself.
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