$f(a)-f(b)$ is rational iff $f(a-b) $ is rational

Prove that the continuous function $f:\mathbb{R} \to \mathbb{R}$ satisfying $f\left(x\right)-f\left(y\right) \in\mathbb{Q} \iff f\left(x-y\right) \in \mathbb{Q}$ is of the form $ f\left(x\right)=ax+b.$

My Attempt.

I tried considering the function $$g\left(x\right)=\frac{f\left(x\right)-f\left(0\right)}{f\left(1\right)-f\left(0\right)}$$ which also satisfies the property $g\left(a\right)-g\left(b\right) \in\mathbb{Q} \iff g\left(a-b\right) \in \mathbb{Q}$,

Now I am trying to prove that this function g is identity function and then I can prove that $f\left(x\right)=\left(f\left(1\right)-f\left(0\right)\right)x+f\left(0\right).$ And I am done.

Also This function has to be identity function because $g\left(0\right)=0$ and $g\left(1\right)=1$.

I tried assuming that the function g is such that $g\left(a\right)\neq a$ for some $a\in \mathbb{R}$. Then by continuity $g\left(x\right)\neq x$ for some $\delta>0$ neighborhood of $a$. But I cannot move further.

Also Using a previously known result, I was able to prove that f must be monotonic. However I do not want to use any other result which is not known and not trivial.

If a function $f $ is continous in $\left[a,b\right]$ and $f\left(a\right)=f\left(b\right)$ then for any $\epsilon >0$ there exists $m,n \in \left[a,b\right] $such that $f\left(m\right)=f\left(n\right)$ and $m-n=\epsilon$.

In this case choose $\epsilon \in \mathbb{R}-\mathbb{Q} $ and get $m,n \in \left[a,b\right] $ such that $f\left(m\right)-f\left(n\right)=0 \in \mathbb{Q}$ but $m-n \in\mathbb{R}-\mathbb{Q}$.

For any $y$, we have either $f(x+y) - f(x) \in \mathbb Q$ for all $x$, or $f(x+y) - f(x) \notin \mathbb Q$ for all $x$, depending on whether or not $f(y) \in \mathbb Q$. But $f(x+y)-f(x)$ is continuous, so by the Intermediate Value Theorem we conclude $f(x+y) - f(x)$ is constant. Thus $$ f(x+y) - f(x) = f(y) - f(0) $$ This says $f(x) - f(0)$ is an additive function. And it's not hard to prove that continuous additive functions are linear.