$C(K,\mathbb{R})$ is complete in the supremum metric for every nonempty compact metric space $K$

By [L1], every $f\in C(K,\mathbb{R})$ is bounded. Thus $C(K,\mathbb{R})$ is a subset of the bounded functions on $K$, and the restriction of the metric in [L2] is a metric on $C(K,\mathbb{R})$.

Let $(f_j)$ be a Cauchy sequence in this supremum metric.

Given a real $\varepsilon>0$, choose $J$ such that $d_\infty(f_m,f_n)<\varepsilon$ for all $m,n\ge J$. Then $|f_m(x)-f_n(x)|\le d_\infty(f_m,f_n)<\varepsilon$ for all such $m,n$ and every $x\in K$, so $(f_j)$ is uniformly Cauchy.

By [L4] there is a function $f:K\to\mathbb{R}$ such that $f_j\to f$ uniformly on $K$.

The function $f$ is continuous by [L5], hence belongs to $C(K,\mathbb{R})$ and is bounded by [L1].

Let $\varepsilon>0$. Uniform convergence gives $J$ such that $|f_j(x)-f(x)|<\varepsilon/2$ for every $j\ge J$ and $x\in K$; hence $d_\infty(f_j,f)\le\varepsilon/2<\varepsilon$, so $f_j\to f$ in the supremum metric.

Every Cauchy sequence in $C(K,\mathbb{R})$ therefore converges in the supremum metric to a member of $C(K,\mathbb{R})$, so the metric space is complete.