Dini's theorem on a closed interval: monotone pointwise convergence of continuous functions to a continuous limit is uniform

Fix a real $\varepsilon>0$. For each $n\in\mathbb{N}$ put $h_n:=f_n-f$ and $U_n:=\{x\in[a,b]:|h_n(x)|<\varepsilon\}$. The function $h_n$ is continuous by [L1], so [L2] makes $U_n$ relatively open in $[a,b]$.

In the nondecreasing case, pointwise convergence forces $f_n(x)\le f(x)$ for every $n,x$, and the errors $f(x)-f_n(x)$ decrease with $n$; in the nonincreasing case it forces $f(x)\le f_n(x)$ and the errors $f_n(x)-f(x)$ decrease. Thus $U_n\subseteq U_{n+1}$ in either case, and pointwise convergence gives $\bigcup_nU_n=[a,b]$.

Let $\mathcal{O}$ be the family of all open sets $O\subseteq\mathbb{R}$ whose trace $O\cap[a,b]$ equals $U_n$ for some $n$. By step 1.1 each $U_n$ has such an open witness, and by step 1.2 the family $\mathcal{O}$ covers $[a,b]$.

By [L3], choose finitely many $O_0,\dots,O_r\in\mathcal{O}$ covering $[a,b]$. By finite choice, choose $n_0,\dots,n_r$ with $O_i\cap[a,b]=U_{n_i}$, and let $N:=\max\{n_0,\dots,n_r\}$.

Since the $U_n$ are increasing, every $U_{n_i}$ is contained in $U_N$; the traces of the $O_i$ cover $[a,b]$, so $U_N=[a,b]$, and then $U_k=[a,b]$ for every $k\ge N$.

Therefore $|f_k(x)-f(x)|<\varepsilon$ for every $k\ge N$ and every $x\in[a,b]$. Since $\varepsilon$ was arbitrary, the convergence is uniform.