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 $$
|f(x)-f(y)| < \epsilon \text{.}
$$
a) There are no functions for which $|f(x_1)-f(x_2)|<-1$ is true. So it is empty set.
b) Let us have $\delta=-1$ and the statement is true. Every function.
c) Every constant function is good. Suppose there are exist $x,y\; x<y,\; f(x)\neq f(y)$. For every positive $\delta: \;x-y<\delta$ but the conclusion can't be true so only constants.
d) Suppose function is not a constant and the conclusion fails immediately. Only constants.
e) Just $x=y$ and no function can hold it. Empty set.
f) Let $f$ have a property: $$\forall x>0\; ax+b\leq f(x)\leq ax+c,\; b\leq c,$$ $$\forall x<0\; Ax+B\leq f(x)\leq Ax+C,\; B\leq C.$$
We obtain for $x, y$ greater than $0$ $$|f(x)-f(y)|\leq|ax+c-ay-b|\leq |a||x-y|+|c-b|$$ and our $\delta$ is greater than $|a|\epsilon+c-b$. For negative $x, y$ it is the same. For $x,y$ of different signs the function is bounded on a closed interval, so the statement is true and we choose the maximum for our bound.
And we can now move $f$ along the $x$-axis - all the conclusions will be the same.
In general I suppose, f) is the case of functions with finite modulus of continuity — it is just a definition. But I do not know if this set has a special name or could it be simplified.
g) First of all, uniformly continuous functions have $$\forall \varepsilon > 0 \; \exists \delta \; |x_1-x_2| < \delta \Rightarrow |f(x_1)-f(x_2)| < \varepsilon$$ or (quite simple) $$\forall \varepsilon > 0 \; \exists \delta \; |x_1-x_2| \leq \delta \Rightarrow |f(x_1)-f(x_2)| \leq \varepsilon$$ or $$\forall \varepsilon > 0 \; \exists \delta \; |f(x_1)-f(x_2)| > \varepsilon \Rightarrow |x_1-x_2| > \delta$$
So it is just the definition of uniformly continous functions.
h) Every bounded satisfies and for unbound one can create an example which will deny the existence of $\epsilon$. Bounded.
i) First of all it should be non-decreasing because $x_1-x_2\leq 0\implies f(x_1)-f(x_2)\leq 0$. And then suppose it is non-decreasing. So for $x>y$ it is $|x-y|>\delta \implies |f(x)-f(y)|<\epsilon$, which is the definition of uniform continuity. For $x\leq y$ it is even more simple. Non-decreasing and uniformly continuous.
Best Answer
Here is a better proof: suppose $T$ is not bounded. Then there exists vectors $x_n$ of norm $1$ such that $\|Tx_n\| >n$ for each $n$. Let $y_n=\frac 1 n x_n+x_0$. Then $y_n \to x_0$ so ( by linearity) $\frac 1 n T(x_n)+T(x_0) \to T(x_0)$. Hence $\frac 1 n \|Tx_n\| \to 0$ which is a contradiction to the choice i=of $x_n$'s.