= Solution
For the first norm-one assertion, use the usual unital <Banach algebra> convention $\|1\|=1$. A <character of an algebra> is a nonzero complex linear multiplicative functional, initially without any continuity assumption. It satisfies $\varphi(1)=1$, since otherwise $\varphi(a)=\varphi(a1)=0$ for every $a$. Moreover, $a-\varphi(a)1$ cannot be invertible: applying $\varphi$ to an inverse identity would give $0=1$. Thus
$$
\varphi(a)\in\sigma_A(a),\qquad |\varphi(a)|\leq r(a)\leq\|a\|.
$$
This proves <automatic continuity of characters> and $\|\varphi\|\leq1$. Evaluation at the normalized identity gives the reverse inequality, so
$$
\boxed{\varphi\text{ is continuous and }\|\varphi\|=1.}
$$
If a convention permits $\|1\|>1$, only $\|\varphi\|\leq1$ follows in general: on $\mathbb C$ with norm $2|z|$, the identity character has norm $1/2$. The later completion argument needs only the upper bound.
If $a$ is invertible, $\varphi(a)\varphi(a^{-1})=1$, so every character is nonzero on $a$. Conversely, if $a$ is not invertible, its principal ideal is proper and lies in a <maximal ideal> $M$ by <Zorn lemma>. The ideal $M$ is closed: if its closure were the whole algebra, it would contain $m$ with $\|1-m\|<1$, making $m$ invertible by the <Neumann series>, which is impossible for a proper ideal. Maximality therefore gives $\overline M=M$. The quotient $A/M$ is a complex Banach division algebra, hence is $\mathbb C$ by the <Gelfand-Mazur theorem>. The quotient map yields a character vanishing at $a$. Consequently
$$
\boxed{a\text{ is invertible}\quad\Longleftrightarrow\quad\varphi(a)\ne0\text{ for every }\varphi\in\Phi_A.}
$$
Now let $A=C(K)$. Its <evaluation characters> give the map $x\mapsto\delta_x\in\Phi_A$. Every character has this form. Indeed, its kernel is a proper ideal. If the functions in that kernel had no common zero, compactness would give finitely many $f_1,\ldots,f_m$ in the kernel with no common zero. Then $h=\sum_j\overline{f_j}f_j$ belongs to the kernel and is everywhere positive, hence is invertible in $C(K)$, a contradiction. At a common zero $x$, the fact that $f-\varphi(f)1$ belongs to the kernel gives $f(x)=\varphi(f)$ for every $f$. Continuous functions distinguish points of the <compact Hausdorff space> $K$, so the map is bijective. It is continuous for the <Gelfand topology>, and the <compact-to-Hausdorff continuous bijection theorem> gives
$$
\boxed{\Phi_{C(K)}\cong K.}
$$
For $K=\varnothing$ the norm comparison is trivial; assume henceforth that $K\ne\varnothing$.
Let $B$ be the completion for the other <algebra norm> $\|\cdot\|_1$. It remains commutative and unital, with $A$ embedded densely. The restriction of any $\psi\in\Phi_B$ is a nonzero character of $A$, since it takes $1$ to $1$, and is automatically continuous for the original <supremum norm>. Thus $R(\psi)=\psi|_A$ maps into $\Phi_A$. It is injective because two continuous characters agreeing on the dense subalgebra $A$ agree on $B$, and it is continuous because each coordinate $R(\psi)(f)=\psi(f)$ is weak-star continuous.
The <character space> $\Phi_B$ is weak-star compact even if its identity was not normalized: the spectral bound puts all its characters in $B_{B^*}$, and the conditions $\psi(1)=1$, $\psi(bc)=\psi(b)\psi(c)$ define a weak-star closed set. Apply <Banach-Alaoglu theorem>. Consequently $R$ is a <homeomorphism> onto a closed subset $L\subseteq K$.
Suppose $x\in U=K\setminus L$. The <compact Hausdorff space> $K$ is normal, so choose an open $V$ with $x\in V\subseteq\overline V\subseteq U$. By the <Urysohn lemma>, choose $g\in C(K)$ with $g(x)=1$ and $g=0$ on $K\setminus V$, and choose $f\in C(K)$ with $f=0$ on $\overline V$ and $f=1$ on $L$. Then $fg=0$. Every character $\psi$ of $B$ restricts to evaluation at a point of $L$, so $\psi(f)=1$. The invertibility criterion proved above makes $f$ invertible in $B$. Multiplying $fg=0$ by $f^{-1}$ in $B$ gives $g=0$, contradicting $g(x)=1$. Therefore $U=\varnothing$ and $R$ is onto $K$.
For each $x\in K$, choose a character $\psi_x\in\Phi_B$ extending $\delta_x$. Its spectral bound gives
$$
|f(x)|=|\psi_x(f)|\leq\|f\|_1.
$$
Taking the supremum over $x$ proves the <minimality of the supremum norm on C(K)>:
$$
\boxed{\|f\|_\infty\leq\|f\|_1\quad(f\in C(K)).}
$$
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