Let $f_{n}\colon\left[0,1\right]\to\mathbb{R}$ be a set of functions such that $f_{n}$ is monotonically increasing for all $n$. Moreover, let's assume that $f_{n}$ converges pointwise to a continuous function $f$. Prove that $f_{n}$ uniformly converges to $f$.
Here is my attempt:
$f\colon$$\left[0,1\right]\to\mathbb{R}$ is a continuous function, therefore from Cantor's Theorem we can say the it is uniformly continuous. That means that there is $\delta>0$ such that for all $x,y\in\left[0,1\right]$ that are close enough $\left|x-y\right|<\delta$ it follows that $\left|f(x)-f(y)\right|<\frac{\varepsilon}{3}$. Now for each $x\in\left[0,1\right]$ let $U_{x}$=$\left(x-\delta,x+\delta\right)$. From the Hiene-Borel Theorem we can conclude that there is a finite set $\left\{ x_{0},\ldots,x_{M}\right\} \subset\left[0,1\right]$ such that $\left[0,1\right]\subset\bigcup_{i=0}^{M}U_{x_{i}}$.
From pointwise convergence we know that for every $0\leq i\leq M$ there is a number $N_{i}$ such that for all $n>N_{i}$ it follows that $\left|f_{n}(x_{i})-f(x_{i})\right|<\frac{\varepsilon}{3}$.
Lets define $N=\max\left\{ N_{0},\ldots,N_{M}\right\} $. All I can conclude for now is that for all $n>N$ and for all $x\in\left[0,1\right]$ there is $0\leq i\leq M$ such that $x\in U_{x_{i}}$ and therefore: $$ \begin{align*} \left|f_{n}(x)-f(x)\right| & \leq\left|f_{n}(x)-f_{n}(x_{i})\right|+\left|f_{n}(x_{i})-f(x_{i})\right|+\left|f(x_{i})-f(x)\right|<\\ & <\left|f_{n}(x)-f_{n}(x_{i})\right|+\frac{2\varepsilon}{3} \end{align*} $$ How can I estimate this $\left|f_{n}(x)-f_{n}(x_{i})\right|$? I know I haven't used the monotony of those $f_{n}$'s yet and it probably has to do something with it. Is this going somewhere?
Choose your points so that $0 = x_0 < x_1 < \dotsc < x_M = 1$. For $x \in \{x_0,x_1,\dotsc, x_M\}$, we have our desired bound. For all other $x$, there is a unique $i$ such that $x_i < x < x_{i+1}$. By the monotonicity of the $f_n$, and the choice of $N$, we have
$$f(x_i) - \frac{\varepsilon}{3} < f_n(x_i) \leqslant f_n(x) \leqslant f_n(x_{i+1}) < f(x_{i+1}) + \frac{\varepsilon}{3}.\tag{1}$$
Using your start, it follows that
$$\lvert f_n(x) - f_n(x_j)\rvert < f(x_{i+1}) + \frac{\varepsilon}{3} - \biggl(f(x_i) - \frac{\varepsilon}{3}\biggr) = f(x_{i+1}) - f(x_i) + \frac{2\varepsilon}{3} < \frac{4\varepsilon}{3}$$
for $j = i$ of $j = i+1$, and thus we obtain $\lvert f_n(x) - f(x)\rvert < 2\varepsilon$ for all $x$ and $n \geqslant N$.
A different argument directly bounding $f_n(x) - f(x)$ shows that for the given $N$ we have $\lvert f_n(x) - f(x)\rvert < \varepsilon$:
As the pointwise limit of monotonic functions, $f$ is monotonic, so
$$f(x_i) \leqslant f(x) \leqslant f(x_{i+1}).\tag{2}$$
Subtracting $(2)$ from $(1)$, we obtain
$$f(x_i) - f(x_{i+1}) - \frac{\varepsilon}{3} < f_n(x) - f(x) < f(x_{i+1}) - f(x_i) + \frac{\varepsilon}{3}.\tag{3}$$
Since $U_{x_i} \cap U_{x_{i+1}} \neq \varnothing$, it follows that
$$0 \leqslant f(x_{i+1}) - f(x_i) < \frac{2\varepsilon}{3},$$
and hence
$$\lvert f_n(x_i) - f_n(x)\rvert < \frac{2\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.$$