= Solution
\b[Invertibility is stable under sufficiently small perturbations, and inversion is continuous.] Let $1$ be the identity of the complex unital <Banach algebra> $A$, with its submultiplicative <algebra norm>. For $\|h\|<1$, completeness makes the <Neumann series>
$$
(1-h)^{-1}=\sum_{n=0}^{\infty}h^n
$$
converge in $A$. Multiplying either side of the partial sums by $1-h$ gives $1-h^{N+1}$, which tends to $1$, proving that the sum is a two-sided inverse.
Fix $a\in G(A)$, the <group of invertible elements of a Banach algebra>. If $\|b-a\|<\|a^{-1}\|^{-1}$, then
$$
b=a(1+a^{-1}(b-a)),\qquad \|a^{-1}(b-a)\|<1,
$$
so the <Neumann series> makes $b$ invertible. This proves that $G(A)$ is open. It also gives a local bound
$$
\|b^{-1}\|\le\frac{\|1\|\,\|a^{-1}\|}{1-\|a^{-1}\|\|b-a\|}.
$$
We retain $\|1\|$ here so the estimate does not silently assume a normalized identity. The inverse identity
$$
b^{-1}-a^{-1}=b^{-1}(a-b)a^{-1}
$$
then implies
$$
\|b^{-1}-a^{-1}\|\le\|b^{-1}\|\|b-a\|\|a^{-1}\|\longrightarrow0\quad(b\to a).
$$
This proves <continuity of inversion in a Banach algebra>.
For a unital <Banach algebra>, the <spectrum of an element> is
$$
\sigma_A(x)=\{\lambda\in\mathbb C:\lambda1-x\notin G(A)\}.
$$
For a nonunital <Banach algebra>, use its <unitization of an algebra> $A^+=A\oplus\mathbb C$ with multiplication and norm
$$
(a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda\mu),\qquad
\|(a,\lambda)\|=\|a\|+|\lambda|.
$$
It is a unital <Banach algebra>, with identity $(0,1)$, and define $\sigma_A(x)=\sigma_{A^+}((x,0))$. In this nonunital convention, $0$ belongs to the <spectrum>: the scalar coordinate of $(x,0)$ is zero, so it cannot be invertible in $A^+$.
\b[The spectrum is nonempty and compact.] It suffices to work in a nonzero unital <Banach algebra>, since <unitization> reduces the other case to this one. The <group of invertible elements of a Banach algebra> is open, so the complement of the <spectrum of an element> is open. For $|\lambda|>\|x\|$, the <Neumann series> gives
$$
R(\lambda)=(\lambda1-x)^{-1}=\frac1\lambda\sum_{n=0}^{\infty}\left(\frac{x}{\lambda}\right)^n,
\qquad
\|R(\lambda)\|\le\frac{\|1\|}{|\lambda|-\|x\|}.
$$
Thus $\sigma_A(x)$ is closed and lies in $\{\lambda:|\lambda|\le\|x\|\}$, hence is compact.
For <nonemptiness of the Banach-algebra spectrum>, suppose that $R$ were defined on all of $\mathbb C$. At any $\lambda_0$, factoring $\lambda1-x=(\lambda_01-x)(1+(\lambda-\lambda_0)R(\lambda_0))$ gives a locally convergent power series
$$
R(\lambda)=\sum_{n=0}^{\infty}(-1)^n(\lambda-\lambda_0)^nR(\lambda_0)^{n+1}.
$$
For every <bounded linear functional> $\ell\in A^*$, the scalar function $\ell(R(\lambda))$ is therefore <entire function>[entire]. The bound at infinity makes it bounded outside a disk, while continuity makes it bounded on the disk. By the <Liouville theorem>, it is constant; since it tends to zero at infinity, it is identically zero. The version of the <Hahn-Banach theorem> used here says that <bounded linear functionals> separate points of a <normed vector space>: if $z\ne0$, there is $\ell$ with $\ell(z)\ne0$. Consequently $R(\lambda)=0$ for every $\lambda$, contradicting $(\lambda1-x)R(\lambda)=1$. This proves the assertion.
\b[Every nonzero complex unital normed division algebra is algebraically and topologically isomorphic to $\mathbb C$.] This is the <Gelfand-Mazur theorem>; completeness is not needed in its statement. Let $D$ be a <normed division algebra>, and take its <completion of a normed space> using the <algebra norm> to obtain a unital <Banach algebra> $\widehat D$. Submultiplicativity extends multiplication continuously to the completion, and the original identity remains its identity.
For $x\in D$, the <spectrum of an element> of $x$ in $\widehat D$ contains some $\lambda$. If $x-\lambda1\ne0$ in $D$, the division-algebra assumption supplies an inverse in $D$, which is still an inverse in $\widehat D$. This contradicts $\lambda\in\sigma_{\widehat D}(x)$. Therefore $x=\lambda1$. The map $\lambda\mapsto\lambda1$ is a bijective complex algebra homomorphism, and
$$
\|\lambda1\|=|\lambda|\|1\|.
$$
It and its inverse are continuous; when $\|1\|=1$, it is an <isometry>.
\b[A complete algebra norm on a function algebra dominates the supremum norm, even before continuity of point evaluations is known.] Let $A$ be the given algebra of functions. Form the same artificial <unitization> $A^+$ even if $A$ already has an identity. For each $t\in K$, the algebraic map
$$
\chi_t:A^+\to\mathbb C,\qquad\chi_t(f,\lambda)=f(t)+\lambda
$$
is a unital multiplicative complex <linear functional>. No continuity has been assumed. Since $\chi_t(f-f(t)1)=0$, the element $f-f(t)1$ cannot be invertible: applying $\chi_t$ to an inverse equation would give $0=1$. Hence $f(t)\in\sigma_{A^+}(f)$, and the <spectrum> bound already proved yields
$$
|f(t)|\le\|f\|\quad(t\in K).
$$
Taking the supremum proves the <supremum bound for a complete function-algebra norm>:
$$
\boxed{\sup_{t\in K}|f(t)|\le\|f\|.}
$$
In particular every $f$ is bounded, and every point evaluation is a <bounded linear functional> of norm at most $1$. This is an instance of <automatic continuity of characters>: a <character of an algebra> on a <Banach algebra> is bounded because its value at any element belongs to that element's <spectrum>.
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