Suppose $f\in F[x]$ is irreducible, $E$ is the splitting field of $f$, and for some $a\in E$ we have $f(a)=f(a+1)=0$. Then $F$ has characteristic $0$.
I'm not sure how to use the last assumption: If $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$, then $f(a)=a^n+\cdots +a_0 = (a+1)^n+\cdots +a_0=0$; I'd have to use the binomial theorem many times to say anything about the $a_n's$ (and I'm not even sure if that would be helpful).
Contradiction: Suppose $F$ has characteristic $p$. Then $f(a)=f(a+kp)$, $k \in \mathbb{Z}$. This is only a contracition if there are more than $n$ distinct elements in $E$ of the form $a+kp$.
May I have a hint?
The statement is wrong. The correct statement is the opposite: Then $F$ has positive characteristic.
Assume $F$ has characteristic $0$. Then $E/F$ is separable, finite and normal, hence a galois extension.
By the assumption, we have some $\sigma \in Gal(E/F)$ with $\sigma(a)=a+1$. This yields $\sigma^n(a)=a+n \neq a$ for all $n \geq 1$. But $\sigma$ is an element of a finite group, hence $\sigma$ has finite order. Contradiction!