Perhaps a clearer answer is ...
Let $f$a addictive continuous function such that $f:\mathbb{R}\rightarrow\mathbb{R}$.
(a) Note that $f$ is linear in $\mathbb{Q}$:
(i) if $q\in \mathbb{Z}$, $f(q)=f(\sum_{i=1}^{q} (1^i))$, for the additivity of $f$, $ f(q)=\sum_{i=1}^{q} f(1^i)=q f(1)=q k $ for some $k\in\mathbb{R}$. So $f$ is linar in $\mathbb{Z}$
(ii) if $q\in \mathbb{Q}\backslash\mathbb{Z}$, $q=\frac{a}{b}$ where $b\neq0$ and $a,b\in \mathbb{Z}$. Note that $f(1)=f\left(\sum_{i=1}^b 1^i/b\right)$, for the additivity of $f$, $ f(1)=\sum_{i=1}^nf( 1^i/n)=nf(1/n)\Rightarrow \frac{1}{b}f(1)=f(1/b)\Rightarrow f(1/b)=k/b $ for some $k\in\mathbb{R}$. So $f$ is linear in $\mathbb{Q}$
(b) Let $x\in \mathbb{R}\backslash\mathbb{Q}$ and $\varepsilon>0$
By the continuity of $f$, there is $\delta>0$ such that $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon$.
For the density of $\mathbb{Q}$ in $\mathbb{R}$, there is $j$ in $(x,y)$, such that $|x-j|<\varepsilon$. $ |f(x)-xf(1)|\leq |f(x) - f(j)| +|f(j) -xf(1)| \leq \varepsilon+|jf(1) -xf(1)| <\varepsilon(1+f(1)) $ as $\varepsilon$ is an arbitrary positive, we have $f(x)=kx$ for any real numbers, that is it $f$ is linear.