Definition: A group $G$ has the minimal condition on subgroups (m.c.s.) if every descending chain $G_1>G_2>...$ in $G$ is finite.
Let $G$ have the m.c.s. and suppose $G$ is not torsion. If $G$ has no proper nontrivial subgroups then $G$ is finite and is torsion. Therefore there is $G_1$ with $G > G_1$ and $G_1 \neq 1$. With the same argument there exists $G_2$ with $G_1>G_2$. And so on. That is, there is an infinite descending chain in $G$, contrary to m.c.s. Hence $G$ is torsion.
Is this proof correct? I'm afraid it's too simple.
No, there's nothing stopping $G_1$ from being finite, which prevents you from constructing the infinite chain. But you can simply take an element that is not torsion and construct an infinite descending chain in the corresponding infinite cyclic subgroup.