Since $f$ is continuous on $(a,b)$, bounded and increasing, there's a unique continous extension of $f$ to $[a,b]$. This works because both limits $f(b) := \lim_{x\to b-}$ and $f(a) = \lim_{x\to a+}$ and are guaranteed to exist since every bounded and increasing (respectively bounded a decreasing) sequence converges. To prove this, simply observe that for a increasing and bounded sequence, all $x_m$ with $m > n$ have to lie within $[x_n,M]$ where $M=\sup_n x_n$ is the upper bound. Add to that the fact that by the very definition of $\sup$, there are $x_n$ arbitrarily close to $M$.
You can then use the fact that continuity on a compact set implies uniform continuity, and you're done. This theorem, btw, isn't hard to prove either (and the proof shows how powerful the compactness property can be). The proof goes like this:
First, recall the if $f$ is continuous then the preimage of an open set, and in particular of an open interval, is open. Thus, for $x \in [a,b]$ all the sets $$
C_x := f^{-1}\left(\left(f(x)-\frac{\epsilon}{2},f(x)+\frac{\epsilon}{2}\right)\right) $$
are open. The crucial property of these $C_x$ is that for all $y \in C_x$ you have $|f(y)-f(x)| < \frac{\epsilon}{2}$ and thus $$
|f(u) - f(v)| = |(f(u) - f(x)) - (f(v)-f(x))|
\leq \underbrace{|f(u)-f(x)|}_{<\frac{\epsilon}{2}}
+ \underbrace{|f(v)-f(x)|}_{<\frac{\epsilon}{2}} < \epsilon
\text{ for all } u,v \in C_x
$$
Now recall that an open set contains an open interval around each of its points. Each $B_x$ thus contains an open interval around $x$, and you may wlog assume that its symmetric around $x$ (just make it smaller if it isn't). Thus, there are $$
\delta_x > 0 \textrm{ such that }
B_x := (x-\frac{\delta_x}{2},x+\frac{\delta_x}{2})
\subset (x-\delta_x,x+\delta_x)
\subset C_x
$$
Note how we made $B_x$ artifically smaller than seems necessary, that will simplify the last stage of the proof. Since $B_x$ contains $x$, the $B_x$ form an open cover of $[a,b]$, i.e. $$
\bigcup_{x\in[a,b]} B_x \supset [a,b] \text{.}
$$
Now we invoke compactness. Behold! Since $[a,b]$ is compact, every covering with open sets contains a finite covering. We can thus pick finitely many $x_i \in [a,b]$ such that we still have $$
\bigcup_{1\leq i \leq n} B_{x_i} \supset [a,b] \text{.}
$$
We're nearly there, all that remains are a few applications of the triangle inequality. Since we're only dealing with finitly many $x_i$ now, we can find the minimum of all their $\delta_{x_i}$. Like in the definition of the $B_x$, we leave ourselves a bit of space to maneuver later, and actually set $$
\delta := \min_{1\leq i \leq n} \frac{\delta_{x_i}}{2} \text{.}
$$
Now pick arbitrary $u,v \in [a,b]$ with $|u-v| < \delta$.
Since our $B_{x_1},\ldots,B_{x_n}$ form a cover of $[a,b]$, there's an $i \in {1,\ldots,n}$ with $u \in B_{x_i}$, and thus $|u-x_i| < \frac{\delta_{x_i}}{2}$. Having been conservative in the definition of $B_x$ and $\delta$ pays off, because we get $$
|v-x_i| = |v-((x_i-u)+u)| = |(v-u)-(x_i-u)|
< \underbrace{|u-v|}_{<\delta\leq\frac{\delta_{x_i}}{2}}
+ \underbrace{|x_i-u|}_{<\frac{\delta_{x_i}}{2}}
< \delta_{x_i} \text{.}
$$
This doesn't imply $y \in B_{x_i}$ (the distance would have to be $\frac{\delta_{x_i}}{2}$ for that), but it does imply $y \in C_{x_i}$!. We thus have $x \in B_{x_i} \subset C_{x_i}$ and $y \in C_{x_i}$, and by definition of $C_x$ (see the remark about the crucial property of $C_x$ above) thus $$
|f(x)-f(y)| < \epsilon \text{.}
$$
There is an infinite number of items in your union, so you can't take the min so easily. You could have $\delta=0$. You have the good idea, but we need to construct the finite union. We will assume U closed, because we can always do the same on the closure of U.
Given $\epsilon>0$ and $x_0=\min U$, we will define $\forall n,\ x_n=\max\left\{
{ x\in \left[x_{n-1}, \infty \right] } \mid
{ \lvert f(x)-f(x_{n-1})\rvert\leq\epsilon }
\right\}$.
Note that $x_n$ can be equal to $\infty$.
$\left(x_n\right)$ is stritly incresing, so converge to
$x_\infty \in \mathbb R_+ \cup \{\infty\}$.
If
$x_\infty \neq \infty$,
then $f$ is not continuous in $x_\infty$, which is not possible.
We have $U $ bounded, and $x_n \rightarrow +\infty$, so $\exists n_0, \forall n>n_0, x_{n} \notin U $. Now $U \subset \bigcup_{i=0}^{n_0} [x_i, x_{i+1}]$ and you have your finite union.
Best Answer
Since $f$ is bounded and monotone increasing $\lim_{x\rightarrow \infty} f(x)=L $ exist. Now suppose $\varepsilon >0 $ and you want to show the uniform continuity requirement for this $\varepsilon$. Choose $R$ such that for $x>R$, $|f(x)-L|< \varepsilon$. Then for $x, y> R$, by monotonicity, $|f( x)-f(y)| < \varepsilon$.
On the other hand, $[0, R+1]$ is compact, so here you get $\delta >0$ such that $|x-y|<\delta$ implies $|f( x)-f(y)| < \varepsilon$ by standard reasoning. Wlog $\delta < 1$, so all of $[0,\infty) $ is covered.