Yes, your proof is entirely correct (except you write "uniform convergence" where you mean "uniform continuity"). Well, to be extremely picky I suppose you should explain what happens when $f'(x_1) = 0$, since then you can't perform the division, but that is a trivial case.
Note that you are actually showing that $f$ is Lipschitz: we say $f: I \rightarrow \mathbb{R}$ is Lipschitz if there is a constant $C$ such that for all $x_1,x_2 \in I$, $|f(x_1)-f(x_2)| \leq C|x_1-x_2|$.
It is helpful to think in terms of the stronger conclusion, because Lipschitz functions have many other nice properties. (For instance, any Lipschitz function is absolutely continuous and thus differentiable almost everywhere. These are more advanced concepts and you may not have encountered them yet, but if you take a graduate level course on real analysis, you certainly will.)
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's an outline of one approach:
$\ \ \ 1)$ Let $\epsilon>0$.
$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/3$ whenever $x\ge N$.
$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N]$.
$\ \ \ 4)$ Choose $ \delta>0$ so that $|f(x)-f(y)|<\epsilon/3$ whenever $|x-y|<\delta $ and $x\in[a,N]$, $y\in[a,N]$.
$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon $ whenever $|x-y|<\delta $ and $x\in[a,\infty)$, $y\in[a,\infty)$
$\ \ \ \ \ \ \ $(consider three cases depending on the relationship between $x$, $y$, and $N$).
A slightly more elegant approach:
$\ \ \ 1)$ Let $\epsilon>0$.
$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/2$ whenever $x\ge N$.
$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N+1]$.
$\ \ \ 4)$ Choose $ 1>\delta>0$ so that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta $, $x\in[a,N+1]$, and
$\ \ \ \ \ \ $$y\in[a,N+1]$.
$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon $ whenever $|x-y|<\delta $ and $x\in[a,\infty)$, $y\in[a,\infty)$
$\ \ \ \ \ \ \ $(consider two cases depending on whether one (or both) of $x$, $y$, exceeds $N+1$).