Suppose $f(x)$ is linear i.e. $f(x + y) = f(x) +f(y) $ and monotone on $[-\infty, +\infty]$, then $f(x) = ax$, a real.

the functional equation shows that $f$ is $\mathbb{Q}$-linear. since for $n$ a positive integer and any real $x$: $$ f(nx) = f(x) + \dots +f(x) = nf(x) $$ with the consequence that $$ f(x) = f(n.\frac{x}{n})=nf(\frac{x}{n}) $$ i.e. $$ f(\frac{x}{n})= \frac1nf(x) $$ hence for positive integers $p,q$ $$ f(\frac{p}{q}.x)=\frac{p}{q}f(x) $$ also $$ f(0)=f(0+0)=f(0)+f(0) $$ implies $f(0)=0$ and $$ f(0)=f(x+(-x))=f(x) +f(-x) $$ then implies $$ f(-x)=-f(x) $$ and so for any rational $r$ we can write $f(r)=rf(1)$

from the functional equation alone that is as far as we can go.

if, in addition, $f$ is monotonic increasing (say), then for any real $x$ we can choose rationals $r_1, r_2$ such that $r_1 \le x \le r_2$. this implies: $$ f(r_1) \le f(x) \le f(r_2) $$ or $$ r_1f(1) \le f(x) \le r_2f(1) $$ using a squeeze this gives us, for any real $x$: $$ f(x)=xf(1) $$ and from this your required conclusion is an easy consequence