Cauchy's equation with monotonicity

We assume $f$ non-decreasing (otherwise replace $f$ by $-f$).

Step 1: algebra. $x=y=0$ gives $f(0)=0$. Then $0=f(x+(-x))=f(x)+f(-x)$, so $f$ is odd. By induction $f(nx)=nf(x)$ for $n\in\mathbb{N}$. For $q=\frac mn$ with $m\in\mathbb{Z}$, $n\ge1$: $nf\left(\frac mn\right)=f(m)=mf(1)$, hence $f(q)=qc$ where $c=f(1)$. Note $c=f(1)\ge f(0)=0$.

Step 2: squeeze. Let $x$ be real and $q_1<x<q_2$ rationals. Since $f$ is non-decreasing,
$$cq_1=f(q_1)\le f(x)\le f(q_2)=cq_2.$$
Letting $q_1,q_2\to x$ (rationals are dense), $f(x)=cx$.

Verification. $f(x)=cx$ is additive.