non-continuous function satisfies $f(x+y)=f(x)+f(y)$
Considering $\mathbb R$ as infinite-dimensional $\mathbb Q$ vector space, any linear map will do. For example, one can extend the function $$f(x)=42a+666b\quad \text{ if } x=a+b\sqrt 2\text{ with }a,b\in \mathbb Q$$ defined on $\mathbb Q[\sqrt 2]$ to all of $\mathbb R$, if one extends the $\mathbb Q$-linearly independent set $\{1,\sqrt 2\}$ to a basis of $\mathbb R$. (This requires the Axiom of Choice, of course)