Attempted proof:
Let $G$ be a group of order $2p$, where $p$ is an odd prime. By Sylow's First Theorem, $G$ contains $p$-subgroups of order $2$ and order $p$. Because both $2$ and $p$ are prime, there exist at most only one Sylow $2$-subgroup and only one Sylow $p$-subgroup. By a corollary, $|Syl_2(G)|$ and $|Syl_p(G)|$ are both normal in $G$. Hence, $G$ is not simple.
Best Answer
For any group $G$, a subgroup $H<G$ of index $2$ is normal (list right and left cosets and draw a conclusion).
If $p\mid |G|$, $G$ has a subgroup with $p$ elements by Cauchy's theorem. But if $|G|=2p$ a subgroup with $p$ elements will have index $2$ by Lagrange's theorem. Thus .....