If $f:[0,\infty)\to [0,\infty)$ and $f(x+y)=f(x)+f(y)$ then prove that $f(x)=ax$ 
Let $\,f:[0,\infty)\to [0,\infty)$ be a function such that $\,f(x+y)=f(x)+f(y),\,$ for all $\,x,y\ge 0$. Prove that $\,f(x)=ax,\,$ for some constant $a$.

My proof :
We have , $\,f(0)=0$. Then , 
$$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(h)}{h}=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=f'(0)=a\text{(constant)}.$$
Then, $\,f(x)=ax+b$. As, $\,f(0)=0$ so $b=0$ and $f(x)=ax.$
Is my proof correct?
 A: In your proof you assume that $f$ is differentiable, which is not given.  
Let me suggest how to obtain the formula of $f$:
Step I. Show that $\,f(px)=p\,f(x),\,$ when $p$ is a positive rational and $x$ a non-negative real. (At first show this for $p$ integer.) We obtain also that, $\,f(0)=0$.
Step II. Observe that $f$ is increasing, since, for $y>x$, we have
$$
f(y)=f(x)+f(y-x)\ge f(x).
$$ 
Step III. 
Since $f$ is increasing, then the limit $\,\lim_{x\to 0^+}f(x)\,$ exists. However
$$
\lim_{x\to 0^+}f(x)=\lim_{n\to\infty}f\Big(\frac{1}{n}\Big)
=\lim_{n\to\infty}\frac{1}{n}\,f(1)=0.
$$
Step IV. Pick an arbitrary $x\in(0,\infty)$, and a decreasing sequence
$\{q_n\}\subset\mathbb Q$ tending to $x$. Then
$$
f(q_n)=q_n\,f(1)
$$
and
$$
x\,f(1)\longleftarrow q_n\,f(1)=f(q_n)=f(x)+f(q_n-x)\longrightarrow f(x),
$$
since $\,\,q_n-x\to 0^+$, and thus $\,\,\lim_{n\to\infty}f(q_n-x)=0$.
Therefore, $\,f(x)=x\,f(1),\,$ for all $x\in\mathbb [0,\infty)$, and hence $\,f'(x)=f(1)$.
A: Note that $f$ is monotonic: if $y>0$, then $f(y)\ge0$ so $f(x+y)=f(x)+f(y)\ge f(x)$.
In particular, the function is continuous over $[0,\infty)$ except for an at most countable set.
You can extend $f$ to $f_e\colon\mathbb{R}\to\mathbb{R}$ by setting $f_e(x)=-f(-x)$, for $x<0$. Show that this function still has the property that $f_e(x+y)=f_e(x)+f_e(y)$. Then continuity at a point implies continuity at $0$.
The result now follows from methods in Overview of basic facts about Cauchy functional equation

Your attempt is not good, I'm afraid: you're assuming differentiability at $0$, which is not among the hypotheses.
A: By induction, $f(nx)=nf(x)$ for an integer $n$.
Now take any real $x$. From
$$\lfloor nx\rfloor\le nx\le\lceil nx\rceil,$$applying the non-decreasing function $f$, we deduce
$$f\left(\lfloor nx\rfloor\right)\le f(nx)\le f\left(\lceil nx\rceil\right).$$ 
By the above induction property,
$$\lfloor nx\rfloor f(1)\le nf(x)\le \lceil nx\rceil f(1),$$ and
$$\frac{\lfloor nx\rfloor}nf(1)\le f(x)\le \frac{\lceil nx\rceil}nf(1).$$
As $n$ can be arbitrarily large, by squeezing
$$f(x)=f(1)x.$$
A: For each $\,m,n\geqslant1\,,$
$f(m)=f\left(n\!\cdot\!\dfrac mn\right)=f\left(\dfrac mn+\dfrac mn+\ldots+\dfrac mn\right)=nf\left(\dfrac mn\right).$
So $\,f\left(\dfrac mn\right)=\dfrac{f(m)}n\,.\;$
Next $\,f(m)=f(1+1+···+1)=mf(1)=ma\,.$
So $\,f\left(\dfrac mn\right)=a\!\cdot\!\dfrac mn\;.$
Also $\,f\left(−\dfrac mn\right)=−f\left(\dfrac mn\right)=a\left(−\dfrac mn\right).$
It follows that for each nonzero $\,x\in\mathbb Q\,,\,f(x)=ax\;$ since $f(0) = 0\,.$
