Q: If $F$ a field then every finite subgroup of $F^*$ is cyclic.
Solution: Suppose $d\mid |G|$ for $G$ subgroup of $F^*$ and $G$ not cyclic. Suppose $A,B$ subgroups of $G$ of order $d$. Then $|A\cup B|>d$ and every $x\in A\cup B$ satisfies $x^d=1$ so $f(x)=x^d-1$ has more that $d$ roots which is a contradiction. So for every $d\mid |G|$ there is at most one subgroup of order $d$ which implies $G$ cyclic.
Is my solution complete or did I miss something?
You should include the word "distinct" when introducing your subgroups ("Suppose $A,B$ are distinct subgroups of order $d$..."). Otherwise, this is good.