Let $f:\Bbb R\to \Bbb R$ be a function that $f(a+b)=f(a)+f(b)$.
It is pretty easy to see that $f(x)=f(1)x$ is a solution, but this solution is not interesting, so I showed that there exists non-continuous solution:
Let look at $\Bbb R$ as vector space over $\Bbb Q$, using the axiom of choice we know that any vector space has a basis, so let $I$ be the basis and so $f(ai)=af(i),a\in\Bbb Q,i\in I$.
For every $x\in\Bbb R$ we have $x=\sum_{i\in I}\lambda_i i,~\lambda_i\in\Bbb Q$ so $f(x)=f\left(\sum_{i\in I}\lambda_i i\right)=\sum_{i\in I}\lambda_i f(i)$
In this case I had to use the basis of $\Bbb R$ over $\Bbb Q$, but without choice I can't guarantee that the basis exists, so do we have to use choice to show existence of non-continuous such function?
