Let $V$ be the infinite dimensional vector space $\mathbb{R}$ over the field $\mathbb{Q}$. Since $1$ and $\sqrt{2}$ are two independent vectors in $V$, we can extend the set $\{1,\sqrt{2}\}$ to a basis $E$ of $V$ (notice that here the axiom of choice is used).

Now we consider a linear function $f\colon V\to\mathbb{R}$ such that $f(1)=1$ while $f(e)=0$ for all $e\in E\setminus\{1\}$. This function is $\mathbb{Q}$-linear (i.e. it is additive on $\mathbb{R}$) but it is not $\mathbb{R}$-linear because $f(\sqrt{2})=0\neq\sqrt{2}f(1)$. ∎