= Solution
For an element $x$ of a unital complex <Banach algebra> $A$, its <spectrum of an element> is
$$
\sigma_A(x)=\{\lambda\in\mathbb C:x-\lambda1_A\text{ is not invertible in }A\}.
$$
If $|\lambda|>\lVert x\rVert$, the <Neumann series>
$$
(\lambda1-x)^{-1}
=\lambda^{-1}\sum_{n\geq0}(x/\lambda)^n
$$
converges, so the spectrum is bounded. If $\lambda_0$ is in the <resolvent of an element>[resolvent set], then
$$
\lambda1-x=(\lambda_01-x)
\left[1+(\lambda-\lambda_0)(\lambda_01-x)^{-1}\right]
$$
is invertible for $\lambda$ sufficiently close to $\lambda_0$, again by a Neumann series. Thus the resolvent is open and the spectrum is closed, hence compact.
If the spectrum were empty, the resolvent $R(\lambda)=(\lambda1-x)^{-1}$ would be an entire Banach-space-valued function and would tend to zero at infinity. For every $f\in A^*$, the scalar entire function $f(R(\lambda))$ would be bounded and therefore constant by <Liouville theorem>. It would be zero because of the limit at infinity. The <Hahn-Banach theorem> would then force $R(\lambda)=0$, contradicting $(\lambda1-x)R(\lambda)=1$. Therefore $\sigma_A(x)$ is nonempty.
Now let $B\subseteq A$ be a closed unital subalgebra containing $x$. Invertibility in $B$ implies invertibility in $A$, so
$$
\sigma_A(x)\subseteq\sigma_B(x).
$$
On each connected component of $\mathbb C\setminus\sigma_A(x)$, either the resolvent belongs to $B$ everywhere or nowhere. Indeed, membership holds on a neighbourhood of any one such point by its local Neumann expansion, and the same argument makes the set of such points relatively closed by taking limits in the closed subalgebra $B$. The unbounded component belongs to the first case because the geometric resolvent series lies in $B$ for large $|\lambda|$. Consequently
$$
\boxed{\sigma_B(x)\text{ is }\sigma_A(x)
\text{ together with a selection of its bounded complementary components}.}
$$
This is the <spectrum in a closed unital subalgebra> theorem.
For the bilateral shift $T$ on $\ell^1(\mathbb Z)$, both $T$ and $T^{-1}$ are isometries. The geometric-series argument applied to $T$ for $|\lambda|>1$ and to $T^{-1}$ for $|\lambda|<1$ gives
$$
\sigma_{B(X)}(T)\subseteq\mathbb T.
$$
For $|\lambda|=1$, normalize the vector given by $x_n=\lambda^{-n}$ on $-N\leq n\leq N$ and zero elsewhere. Only its two boundary coordinates contribute to $(T-\lambda I)x$, so the norm of that image tends to zero. Thus $T-\lambda I$ is not bounded below and cannot be invertible. Hence
$$
\boxed{\sigma_{B(X)}(T)=\mathbb T.}
$$
The algebra generated by $T$ is the norm closure of polynomials in nonnegative powers of $T$. It does not contain $T^{-1}$: every $p(T)e_0$ is supported in nonnegative coordinates, whereas $T^{-1}e_0=e_{-1}$. Therefore $0\in\sigma_{\mathcal A}(T)$. The preceding hole theorem then fills the unique bounded component of $\mathbb C\setminus\mathbb T$, giving
$$
\boxed{\sigma_{\mathcal A}(T)=\overline{\mathbb D}.}
$$
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