Let $\mathcal{A}$ be an abelian Banach algebra, and let $h : \mathcal{A} \to \mathbb{C}$ be a homomorphism.
The standard proof(which I have now seen in at least 2 books) for the theorem given in the title is as follows.
Let $a \in \mathcal{A}$. Note that $h(a - h(a)e) = 0$, so it follows that the $a - h(a)$ belongs to the kernel of $h$ which is a maximal ideal. Maximal ideals cannot contain invertible elements, and hence $a - h(a) e$ is not invertible, and $h(a) \in \sigma(a)$.
If $\rho(a)$ is the spectral radius of the element $a$, then recall that for Banach algebras we have $$\rho(a) = \lim_{n \to \infty} ||a^n||^\frac{1}{n} \leq || a || \ .$$ Combining these two theorems, for any $a \in \mathcal{A}$, we have $$ |h(a)| \leq \rho(a) \leq || a || \ .$$
I think I have an alternative proof which is more elementary.
First, we note that these homomorphisms are necessarily continuous since their kernels are maximal ideals, and maximal ideals are closed, so the continuity of $h$ follows by studying it on the quotient space $\mathcal{A} / \operatorname{ker}(h)$.
Now, let $h$ be some homomorphism as above, and assume that $$ || h || = \sup_{a \in \mathcal{A}, || a || = 1} | h(a)| > 1 \ .$$ This implies that there exists some $a$ in the unit sphere such that $|h(a)| > 1$. Since $h(a)$ is a complex number with modulus greater than $1$, we can write it in the form $$ h(a) = (1 + \varepsilon) e^{i \theta}$$ for some $\varepsilon > 0$ and $\theta \in [0, 2\pi)$. Define $$ a' = \frac{a}{1 + \varepsilon} \ .$$ Since $|| a || = 1$, we have $$ || a' || = \frac{1}{1 + \varepsilon} < 1 \ ,$$ and $$ |h(a')| = \frac{1 + \varepsilon}{1 + \varepsilon} = 1 \ .$$ Finally, define $$x_n = (a')^n$$ and note that $$|| x_n || \leq || (a')||^n = \left( \frac{1}{1 + \varepsilon}\right)^n$$ and $$|h(x_n)| = |h((a')^n)| = |h(a')|^n = 1 \ .$$ We see that $x_n$ is a sequence in $\mathcal{A}$ which converges to $0$, and thus by continuity $h(x_n) \to 0$, in particular the norm of $h(x_n)$ should converge to $0$, but clearly it doesn't. This is a contradiction, so we must have $$|| h || \leq 1 \ .$$ I have read my proof through a few times, and I cannot find any errors. My approach seems simple enough that I am surprised I have not seen it elsewhere. Any comments would be appreciated, I would also be interested in seeing some other proofs of this theorem, in particular, I would be interested in proofs which ignore using the spectral radius.