Continuous functions form a commutative Banach algebra

The supremum norm is finite on every $f$: $\mathrm{\mid }f\mathrm{\mid }$ is continuous and real-valued on the nonempty compact $K$, so it attains a maximum by [L1]; the pointwise operations make $C(K,\mathbb{C})$ a commutative associative complex algebra with unit the constant function $1$, and $\mathrm{\parallel }fg{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ holds because $\mathrm{\mid }f(x)g(x)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $x$ while $\mathrm{\mid }1(x)\mathrm{\mid }=1$ gives $\mathrm{\parallel }1{\mathrm{\parallel }}_{\mathrm{\infty }}=1$.

Completeness: a $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\mathrm{\infty }}$-Cauchy sequence $(f_n)$ is uniformly Cauchy, so by [L2] it converges uniformly to a continuous $f$; uniform convergence is convergence in the supremum norm, so $C(K,\mathbb{C})$ is complete.

Spectral inclusion: if $\lambda \mathrm{\notin }f[K]$ then $m:={\inf }_{x\in K}\mathrm{\mid }\lambda -f(x)\mathrm{\mid }>0$: the function $x\mapsto \mathrm{\mid }\lambda -f(x)\mathrm{\mid }$ is continuous on the nonempty compact $K$ and attains its minimum $m$ by [L1], and $m=0$ would mean $\lambda =f(x)$ for some $x$. Hence $1\mathrm{/}(\lambda -f)$ is a bounded continuous function with $\mathrm{\parallel }1\mathrm{/}(\lambda -f){\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}m$, and it is a two-sided inverse of $\lambda 1-f$; so $\lambda \in \rho (f)$.

Spectral equality: conversely, if $\lambda =f(x_0)$ for some $x_0\in K$ and $g$ were an inverse of $\lambda 1-f$, then evaluating the identity $g\cdot (\lambda 1-f)=1$ at $x_0$ would give $g(x_0)\cdot 0=1$, impossible; hence $\lambda \in \sigma (f)$. Combined with [step 2.2], $\sigma (f)=f[K]$.