Let $f$ be entire function satisfying $f(z)=f(z+\xi t)$ for all $t\in\mathbb{R}$, some $\xi\in\mathbb{C}\neq 0$ and all $z\in\mathbb{C}$. I would like to show that $f$ is constant.
Since $f(z+\xi t)-f(z)=0$ for all $z\in\mathbb{C}$, we have that by the holomorphicity of $z$, $$0=\xi\lim_{t\xi \to 0}\frac{f(z+\xi t)-f(z)}{\xi t}=\xi f'(z)$$ Thus, we must have $f'(z)=0$ everywhere and $f$ is constant.
Is the proof correct and is there another proof of this result?
Yes, it is fine. Even less would have made $f$ constant. For example, the condition is not needed for all $z$, just one $z_0$ would be enough. This is because the function $g(z)=f(z_0+z)-f(z_0)$ is zero on a set, all $\xi t$, that accumulates inside where $g$ is analytic. Therefore $g(z)$ is identically zero.