An exercise on a function $f$ satisfying $f(x+y)=f(x)+f(y)$ I tried to solve the following exercise found in Foundations Of Modern Analysis - J. Dieudonné vol 1, section 4.2.

My solution for the first question was :
Suppose we have some $M\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $f\left( x\right) \leq M$ for all $x\in \left] a,b\right[ $. Now
since $a<a+b-x<b$ for any $x\in \left] a,b\right[ $, then we must have $f\left(
a+b-x\right) \leq M$ so $f\left( a+b\right) -f\left( x\right) \leq M$ and
then $f\left( x\right) \geq f\left( a+b\right) -M$.
I want to know how to use hints given by the Author, because I think my very
simple argument is false.
 A: Assume $f$ is bouned $\bf above$ in $(a,b)$.
(1) Assume $a<x<c<y<b$ with $\frac{y-c}{c-x}=\frac{p}{q}\in\mathbb Q$.
Denote $$M_1=\min\{f(x),f(y)\}.$$
Then
$$qy-qc=pc-px\implies (p+q)c=px+qy\implies f(c)=\frac{p}{p+q}f(x)+\frac{q}{p+q}f(y).$$
We have
$$f(c)\geq M_1,\quad c\in\Lambda=\{x+\lambda (y-x)\mid \lambda \in \mathbb Q\}.$$
(2) By translation, we have $f(w)\leq M_2$ for $w\in [0,y-x]$.
(3) Now consider $x<z<y$. Then we can choose a $\Delta z\in [0,y-x]$ such that $z+\Delta z=c\in \Lambda$. We have
$$f(z)=f(c)-f(\Delta z)\geq M_1-M_2,$$
that is $f$ is bounded $\bf below$ in $[x,y]$.
(4) By translation, $f$ is $\bf bounded $ in any compact interval.
(5) To show $f$ is continuous at $z$, we need to show
$$\lim_{\Delta z\to 0} f(z+\Delta z)=f(z)\iff \lim_{\Delta z\to 0} f(\Delta z)=0.$$
Let $M=\sup_{x\in [-1,1]}|f(x)|$. Then $|f(\Delta z)|\leq \frac{M}{n}$ if $|\Delta z|\leq \frac{1}{n}$.
(6) Now let $f(1)=c$, note that $f(0)=0$, we have $f(\frac{p}{q})=\frac{p}{q}\cdot c$. By continuity, $$f(x)=\lim_{z\to x,~z\in \mathbb Q}f(z)=\lim_{z\to x,~z\in \mathbb Q} z\cdot c=x\cdot c.$$
