If $f:[0,\infty)\to [0,\infty)$ and $f(x+y)=f(x)+f(y)$ then prove that $f(x)=ax$

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 $\,f'(x)=f(1)$.

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.

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

$$f(x)=f(1)x.$$