Let $K$ be a finite Galois extension of $F$ of characteristic $0$. Show that if Gal$\left(K/F\right)$ is a non-trivial $2$-group, then there is a quadratic extension of $F$ contained in $K$.
If the Galois group $G$ is a $2$-group then each element must have an order a power of $2$. Hence, I have to show that $G$ has element of order $2$. However, I don't know where to start on this. Any help would be much appreciated.
A group of even order has an element of order $2$ by Cauchy's theorem. However, this is not what you need. By the fundamental theorem of Galois theory (of course we use the fact that it is a finite Galois extension) there is one to one and onto correspondence between the set of fields between $F$ and $K$ and the set of subgroups of $\operatorname{Gal}(K/F)$. If $H\leq \operatorname{Gal}(K/F)$ corresponds to the field $M$ then $[ \operatorname{Gal}(K/F): H]=[M:F]$. Note that what I wrote on the right side of the equality is an extension degree while on the left side it is an index of a subgroup. So since you want to prove there is a field $F\subseteq M\subseteq K$ such that $[M:F]=2$ then what you actually need is to prove that $G:= \operatorname{Gal}(K/F)$ has a subgroup of index $2$, not of order $2$.
Well, this easily follows from a theorem in group theory which says that any $p$-group of size $p^n$ has subgroups of order $1,p,p^2,...p^{n-1}$. It was proved a few times on this forum, for example here: $p$-groups have normal subgroups of each order
