Clarification: this is not Dini's theorem.
The title is highly succinct but here is the task in full detail:
Let $[a,b]\in \mathbb{R}$ where $a<b$ are reals.
Each $f_n: [a,b] \to \mathbb{R}$ is an increasing function. (Namely $x<y \implies f_n(x) \leq f_n(y)$ for all reals $x,y$ and all natural $n$).
Suppose $f_n$ converges pointwise to $f$. Show
(a) $f$ is also increasing,
(b) if $f$ is continuous then the convergence is uniform.
(a) was trivial, but I have been grappling with (b) recently. Today I think I managed to conjure up a passable solution, albeit a somewhat gory one. I present it below.
I am curious to know whether it has any logical gaps and if it could be improved upon and condensed.
Attempt at (b):
$f$ is continuous on $[a,b]$ and thus uniformly continuous. Let $\epsilon > 0$ be given, upon which there is a $\delta(\epsilon) > 0$ such that for each $x \in [a,b]: f\big( B(x,\delta) \big) \subset B(f(x), \epsilon) $.
To show uniform convergence we wish to exhibit a natural $N$, depending solely on $\epsilon > 0$, such that $n \geq N$ implies $|f_n (x) - f(x)| < \epsilon$ for each $x\in [a,b]$.
Take any $x \in [a,b]$ and draw $B(x,\frac \delta 2) \cap [a,b]$. The strategy next is to find a natural $N$ that works for all points in $B(x, \frac \delta 2)$. After that we will cover $[a,b]$ with open balls $B(x,\frac \delta 2)$ and extract a finite subcover, upon which we should obtain a natural number that works everywhere on $[a,b]$, thus affirming the convergence is uniform.
We know $f$ is increasing and $f$ maps $B(x,\frac \delta 2)$ into $B(f(x), \epsilon)$. Therefore $f(x) - \epsilon < f(x-\frac \delta 2) \leq f(x) \leq f(x+\frac \delta 2) < f(x) +\epsilon $.
Pointwise convergence is ensured at $x+\frac \delta 2$. Thus there is a natural $N(x+\frac \delta 2, \epsilon)$ such that $n \geq N(x+\frac \delta 2, \epsilon)$ implies $f_n(x+\frac \delta 2) < f(x+\frac \delta 2) + \epsilon < f(x) + 2 \epsilon$.
But $f_n$ is increasing, so whenever $t < x+ \frac \delta 2$ and $ t\in[a,b]$, we have $f_n (t) \leq f_n (x+\frac \delta 2) <f(x) + 2\epsilon $. Of course we also have $f(t) < f(x) + \epsilon$.
Similarly we have a natural $N(x-\frac \delta 2,\epsilon)$ such that whenever $t > x - \frac \delta 2$ and $t\in [a,b]$ and $n \geq N(x-\frac \delta 2, \epsilon)$, we have $f_n(t) \geq f_n (x-\frac \delta 2) > f(x) - 2\epsilon$. Again we also have $f(t) > f(x) - \epsilon$.
So if $n \geq \max\{ N(x-\frac \delta 2, \epsilon), N(x+\frac \delta 2,\epsilon) \}$ and if $t\in B(x,\frac \delta 2) \cap [a,b]$ then we have $|f(t) -f(x) |< \epsilon$ and $|f_n(t) - f(x)|<2\epsilon$. Thus we have $|f_n(t) - f(t)| < 3 \epsilon$.
Thus if we go far enough beyond $N(x,\epsilon) := \max\{ N(x-\frac \delta 2, \epsilon), N(x+\frac \delta 2,\epsilon) \}$ then $f_n$ is within $3\epsilon$ of $f$, everywhere in the ball $B(x,\delta/2)$.
Now cover $[a,b]$ with $\bigcup_{x\in [a,b]} B(x,\frac \delta 2)$. Compactness furnishes a finite subcover, so say $[a,b] \subset B(x_1, \frac \delta 2) \cup B(x_2, \frac \delta 2)$. We showed earlier there are naturals $N(x,\epsilon)$ making the convergence "uniform" in $B(x, \frac \delta 2)$. The finite subcover therefore furnishes naturals $N(x_1, \epsilon)$ and $N(x_2, \epsilon)$.
Simply take $N(\epsilon) \geq \max\{ N(x_1, \epsilon), N(x_2, \epsilon) \}$. Then if $t\in [a,b]$, $t$ is in some ball $B(x,\frac \delta 2)$ and since $N(\epsilon)$ is large enough, $n \geq N(\epsilon)$ will ensure $|f_n(t) -f (t) | < 3 \epsilon$. Thus the convergence is uniform on $[a,b]$. We are done. $\blacksquare$.
The proof can be made much shorter and I believe more transparent.
By the uniform continuity of $f,$ for $\varepsilon>0$ there exists $m$ such that $|f(x_k)-f(x_{k-1}|<\varepsilon/2,$ where $x_k=a+{k\over m}(b-a),$ $k=0,1,\ldots,m.$ By the pointwise convergence there exists $N$ such that for any $n>N$ and $l$ we have $|f_n(x_l)-f(x_l)|<\varepsilon/2.$ Next let $x_{k-1}\le x\le x_k.$ Then $$f(x_{k-1})-{\varepsilon\over 2}\le f_n(x_{k-1})\le f_n(x)\le f_n(x_k)\le f(x_k)+{\varepsilon\over 2}$$ $$f(x_{k-1})\le f(x)\le f(x_k)$$ Therefore for $n>N$ we get $$|f_n(x)-f(x)|< \varepsilon$$