Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is additive ($f(x+y)=f(x)+f(y)$) and monotonic on a set $D\subset\mathbb{R}$ such that $|D|>1$, $0\in D$ and $-a\in D$ whenever $a\in D$. Assume nothing about the behavior of $f$ in $\mathbb{R}\setminus D$.
Is it true that, for all $x\in D$, $f(x)=\alpha x$ for some $\alpha\neq 0$?
The answer is "No!" considering axiom of choice which is sufficient to prove that there is a Hamel basis.
Let $ H $ be a Hamel basis containing, say $ 1 $ and $ \sqrt 2 $. Define $ f $ such that $ f ( 1 ) = 1 $, $ f \left( \sqrt 2 \right) = 2 $ and $ f ( a ) = 0 $ for each $ a \in H \setminus \left\{ 1 , \sqrt 2 \right\} $. Then taking $ D = \left\{ - \sqrt 2 , -1 , 0 , 1 , \sqrt 2 \right\} $, $ f $ is increasing on $ D $ but there is no $ \alpha \in \mathbb R $ such that $ f ( x ) = \alpha x $ for all $ x \in D $.