Dini's Theorem Proof on the Reals Dini's Theorem states that:

Let $K$ be a compact metric space.  Let $f:K→\mathbb R$ be a continuous function and $f_n:K→ \mathbb R,n∈\mathbb N$, be a sequence of continuous functions.  If ${f_n}$ converges pointwise to $f$ and if $f_n(x)≥f_{n+1}(x)$ for all $x∈K$ and all $n∈\mathbb N$ then ${f_n}$ converges uniformly to $f$.

It is usually proven using a finite open subcover of $K$, like here. My question: is there a more elementary proof if  $K$ is a compact in $\mathbb R$, namely a closed and bounded interval $[a,b]$?
The idea would still be to prove that  $\lim_{x\to 0}\sup|{f_n-f}|=0$. We still have that $f-f_n$ is decreasing and converges pointwise to $0$. It's also continuous on $[a, b]$, so its $\sup$   is actually a maximum. This should be enough to say that $\lim_{x\to 0}\max|{f_n-f}|=0$, but I'm missing the steps: how do I use the fact that $f-f_n$ is decreasing to show that its maximum converges to $0$?
 A: This proof on $\mathbb{R}$ was given by the solutions manual for Spivak's Calculus, and it follows as such:
First we prove for the case when $f_n$ converges pointwise to $ f(x) = 0$ on $[a, b]$ and $f_n(x) >= f_{n+1}(x)$.
Suppose that $f_n$ doesn't converge to $0$ uniformly, thus for some $\epsilon \gt 0$ there always will be an $ x \in [a, b]$ such that $ f_n(x)  \ge \epsilon $ holds for arbitrarily large $n \in \mathbb{N}$. Therefore we can construct a sequence of $n_1 < n_2 < n_3 < \dots $ indices with a corresponding $\{x_n\} $ sequence such that $$  f_{n_i}(x_i)  \ge \epsilon \tag{1}$$
Since $\{x_n\}$ sequence is bounded, it has a convergent subsequence $\{ 
x_{k_n}\} $, and let $ l = \lim_{n \to \infty} x_{k_n}$. $ 0 = f(l) = \lim_{n \to \infty} f_n(l) $, so there will be some $N > 0$ such that $f_n(l) \lt \epsilon $ holds for $n > N$. Furthermore each $f_n$ is continuous, so $f_n(y) < \epsilon$ will hold for all $y$ close enough to $l$, so in particular also for some $x_{k_i}$. By choosing $i$ large enough, $n_{k_i} > n$ also holds, thus $ f_{n_{k_i}}(x_{k_i}) \lt \epsilon$. But this contradicts the choice of $\{x_n\}$ in $(1)$.
Now if $f_n$ converges to some continuous f instead of 0, we can just apply the first part to $f_n - f$.
