As a comment on your attempt, note that $R_{n,b,f}$ is not a polynomial. However, by hypothesis it can be written $R_{n,b,f}(x)=(x-b)^n r_{n,b,f}(x)$ where $\lim \limits_{x\to b} r_{n,b,f}(x)=0$.
The proof of (4) involves the corrected versions of both (c) and the remark at the end of (c). It doesn't require $a=0$ or $b=0$ but does assume $b=g(a)$.
Substitute $g(x)$ into $R_{n,b,f}(x) = (x-b)^n r_{n,b,f}(x)$ to obtain
$$B:=R_{n,b,f}(g(x)) = [g(x)-b]^n r_{n,b,f}(g(x)).$$ We can write
$$
[g(x)-b]^n=\big[P_{n,a,g}(x) + R_{n,a,g}(x)-b\big]^n = p\big(q(x) + R_{n,a,g}(x)\big)
$$ where $p$ and $q$ are the polynomials
$$p(x):=[x-b]^n,\qquad q(x):=P_{n,a,g}(x).$$
Since $R_{n,a,g}(x)/(x-a)^n\to0$ as $x\to a$, we can apply (c) to get
$$
[g(x)-b]^n=p(q(x)) + \overline R(x)
$$ where $\lim\limits_{x\to a}\frac{\overline R(x)}{(x-a)^n}=0$. Note now that $q$ is a polynomial in $x-a$, while $p$ is a polynomial in $x-b$ with $b:=g(a)=P_{n,a,g}(a)=q(a)$. So by the remark, $p(q)$ is a polynomial in $x-a$ with degree at least $n$.
Divide $B$ by $(x-a)^n$:
$$\frac B{(x-a)^n} = \left[
\frac{p(q(x))}{(x-a)^n} + \frac{\overline R(x)}{(x-a)^n}
\right]r_{n,b,f}(g(x))
$$
As $x\to a$, the first term in square brackets tends to a constant, the second term tends to zero, and $r_{n,b,f}(g(x))$ tends to zero because $g(x)$ tends to $g(a)=b$. This proves (4).
As for (5), there's no need to invoke (c) again. The argument establishes two Taylor polynomials of degree $n$ for $f\circ g$ at $a$, both of whose remainders, when divided by $(x-a)^n$, tend to zero as $x\to a$, hence the two polynomials are equal.
There is some room for interpretation in answering your question.
Let me start by showing the following. Assume $f$ is continuous and it satisfies $\lim_{x\to\infty} \frac{1}{x}\int_x^{x+1} f(t)dt=\ell$. This is NOT sufficient to argue that $\lim_{x\to\infty} \frac{f(x)}{x}=\ell$.
Indeed, consider the function $f(x)=x\sin{2\pi x}$. It is not difficult to see by integrating by parts that $\int f(x)dx= \frac{1}{4\pi^2}(\sin{2\pi x}-2\pi x\cos{2\pi x})+C$, so $\int_x^{x+1} f(t)dt=\frac{1}{2\pi} \cos{2\pi x}$. Therefore $\ell=0$ in this case. However, $\lim_{x\to\infty} \frac{f(x)}{x}=\lim_{x\to\infty} \sin{2\pi x}$ does not exist.
So if your approach only uses the assumption that $\lim_{x\to\infty} |f(x+1)-f(x)|=\ell$ to get the limit for the integral, it will not work. Note that with my example, this limit is again exactly the limit of $\sin{2\pi x}$ which does not exist.
Therefore, if you want to take this approach, the question boils down to being able to use the existence of the limit of $|f(x+1)-f(x)|$ again after the integral limit in a way that is different from the suggested textbook proof. If you cannot do that, then probably this is a dead end.
Best Answer
Given that $a_n:=\inf f[n,n+1], b_n=\sup f[n,n+1] .$
Well-definedness: $a_n,b_n$ are well defined since $f$ is continuous and $[n,n+1]$ is a compact set (this is same as closed and bounded in $\mathbb R$).
To prove: $\lim_n (a_{n+1}-a_n)=l=\lim_n (b_{n+1}-b_n),$ which by definition of limit is equivalent to proving that "Given any $\epsilon>0$, there exists $N\in \mathbb N$ such that $n\ge N\implies -\epsilon\lt a_{n+1}-a_n-l\lt \epsilon.$"
To achieve this, first fix an $\epsilon>0$.
Let's break down rest of the solution in steps:-
Step (1): (Choosing large $x$) Since $\lim_{x\to \infty}(f(x+1)-f(x))=l$, there exists $N\in \mathbb N$ such that $x\ge N\implies -\epsilon/2\lt f(x+1)- f(x)-l\lt \epsilon/2.$
Step(2):(Estimating $a_{n+1}-a_n$ from above) Note that for all $n\ge 2N,$
By definition of infimum, there exists $x_n\in [n,n+1]$ such that $a_n+\epsilon/2\gt f(x_n).$ It follows that $$\begin{align} a_{n+1}-a_n-l \le &a_{n+1}-f(x_n)-l+\epsilon/2\\ \le& f(x_n+1)-f(x_n)-l+\epsilon/2\\ \overbrace{\le }^{\text{ By step $(1)$}}& \epsilon/2+\epsilon/2=\epsilon \end{align}$$
Step (3):(Estimating $a_{n+1}-a_n$ from below) Note that for all $n\ge 2N,$
By definition of infimum, there exists $y_n\in [n+1,n+2]$ such that $a_{n+1}+\epsilon/2\gt f(y_n).$ It follows that $$\begin{align} a_{n+1}-a_n-l \ge & f(y_n)-a_n-l-\epsilon/2\\ \ge& f(y_n)-f(y_n-1)-l-\epsilon/2\\ \overbrace{\ge }^{\text{ By step $(1)$ again}}& -\epsilon/2-\epsilon/2=-\epsilon \end{align}$$
It follows by step (2) and (3) that: Given any $\epsilon>0$, there exists $N\in \mathbb N$ such that $n\ge 2N\implies -\epsilon\lt a_{n+1}-a_n-l\lt \epsilon.$ This proves the statement for $(a_{n+1}-a_n)$. The statement is similar for $(b_{n+1}-b_n)$, which I leave up to you.