By Dini's theorem, I mean
$E$ is a compact metric space.
$f: E \to \mathbb{R}$ and $f_n: E \to \mathbb{R} $ are all continuous functions. Moreover $f_n \to f$ pointwise.
Suppose for each $p \in E$ that $f_n (p) \leq f_{n+1} (p)$ for all naturals $n\geq 1$.
Prove $f_n \rightrightarrows f $, namely, the convergence is uniform.
I have seen Rudin's proof and it is neat and concise. But I wanted to attempt a proof of my own, be it as gory and verbose as it may.
I think I have managed to construct an argument, and wanted ask for feedback and know whether I have overlooked anything.
Attempt:
Given $\epsilon >0 $, we want a $N(\epsilon)$ such that $n \geq N(\epsilon)$ implies $|f_n(t)- f(t)| < \epsilon$ for each $t \in E$.
My strategy is to find, for each $p \in E$, a $N(p,\epsilon)$ that suffices in some ball at $p$. Subsequently we cover $E$ with these balls and extract a finite subcover. This will furnish a finite number of naturals $N(p,\epsilon)$, and we can simply take the maximum which will work over the whole space $E$.
So, given $\epsilon > 0$, take any $p \in E$.
As $f_n \to f$ pointwise, there is $ N:= N(p,\epsilon)$ such that $n\geq N$ implies $|f_n(p) - f(p)| < \epsilon$.
$f$ is continuous at $p$ so there is $\delta > 0$ such that if $t \in B(p, \delta)$ then $|f(t) - f(p)| < \epsilon$.
The particular function $f_N$ is continuous at $p$ so there is $\delta_N >0$ such that if $t \in B(p, \delta_N)$ then $|f_N(t) - f_N(p)| < \epsilon$.
If $t \in B(p,\delta) \cap B(p,\delta_N) $ then
$|f_N (t) - f(t)| \leq |f_N (t) - f(p)| + |f(p)-f(t)| \leq |f_N(t) - f_N(p)| + |f_N (p) - f(p)| + |f(p) - f(t)| < 3\epsilon $
We have$f_n(t) \leq f(t)$ for all natural $n$. (If not then for some $k$ we have $f(t) < f_k(t)$ and thus $f(t) < f_k(t) \leq f_{k+1} (t)\leq f_{k+2} (t)\leq \ldots$, which would violate pointwise convergence.)
So if $t \in B(p,\delta) \cap B(p,\delta_N) $ then for each $n \geq N$ we have $f_N (t) \leq f_n (t) \leq f(t)$. That is
$-3 \epsilon <f_N (t) - f(t) \leq f_n (t) - f(t) \leq 0 < 3\epsilon \implies |f_n(t) - f(t) | < 3 \epsilon $
Hence $N(p,\epsilon)$ suffices for all points in the ball $B(p,\delta) \cap B(p,\delta_N)$. We can now, as intimated in our strategy outline, simply cover $E$ with these balls, extract a finite subcover, and from here obtain an $N(\epsilon)$ that works on the entire $E$. Hence $f_n \rightrightarrows f$. We are done. $\blacksquare$.