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 $$
|f(x)-f(y)| < \epsilon \text{.}
$$
Best Answer
I came across this problem when I reading this undergrad analysis book, and I think this problem is a little bit challenging unless I was overdoing it. Here is my proof:
First we have $A=\bigcup_{\delta>0} A(\delta)$ is bounded above by $b$ by the construction of $A(\delta)$, and it's nonempty because it contains $a$. So there exists $c=l.u.b.(A)$, of course $c\leq b$ since $b$ is an upper bound.
Some fact: (1) if $u \in A(\delta)\subset A$, then for any $v\in[a,u)$, $v$ is also in the same $A(\delta)$, therefore in $A$. This is just saying that if there is any $u\in A$, then $[a,u]\subset A$. Proof: play with the definition.(omitted)
(2)For fixed $\epsilon$, if $s\in A$ where A is induced by $\epsilon$, then $s\in A$ where A is induced by $\frac{\epsilon}{2}$. Proof: rescale $\epsilon$.
Since function $f$ is continuous at $c$, for any $\epsilon >0$,there exists $\delta_0>0$ such that for any $y$ that satisfies $|y-c|\leq\delta_0$ we have $|f(c)-f(y)|\leq \epsilon$. Now we pick this very $\delta_0$, since $c$ is the l.u.b.(A) so there exists $s\in A$ such that $c-\delta_0\leq s\leq c$, since $s\in A=\bigcup_{\delta>0}A(\delta)$, so there exists $\delta_1$ such that $s\in A(\delta_1)$. Now we want to show that $c$ is also in $A$:
Fact (1) tells us $[a,s]\subset A(\delta_1)$, let $\delta=min\{\delta_0, \delta_1\}$. Then for any $x,t \in [a, c]$ with $|x-t|<\delta$,
case1: If $x,t$ are both in $[a,s]$, then $|f(x)-f(t)|<\epsilon$ because $f$ is uniformly continuous on $[a,s]$.
case2: If $x,t$ are both in $[s, c]$, then $|f(x)-f(t)|\leq |f(x)-f(c)|+|f(c)-f(t)|<\epsilon + \epsilon=2\epsilon$.
case3: If one is in $[a,s]$ and the other is in $[s,c]$. WLOG, let $x\in [a,s]$ and $t\in [s,c]$, then $|f(x)-f(t)|\leq |f(x)-f(s)|+|f(s)-f(c)|+|f(c)-f(t)|\leq \epsilon + \epsilon +\epsilon =3\epsilon$.
Anyway, we can use fact (2) to rescale and get a uniform $\epsilon$. So we've shown $c$ is also in $A$. Now if we assume $c<b$, then we can repeat the above argument to show $c+\delta_0$ is also is also in $A$ (use $c$ as the intermediate point), contradicting the fact the $c$ is the l.u.b.(A). So the only possible way is that $c=b$ and $c+\delta_0$ is out of the bound of domain $[a,b]$. Therefore $c=b\in A$, done.
P.S.: I've carefully checked my reasoning and it should be right up to some detail work, e.g. rescaling issue. And I'm pretty sure there is a better way to put all these together nicely, but I'm out of time so...