Torsion and Torsion-Free Quotient Group

If $T$ is the torsion subgroup of an abelian group $G$, then $G/T$ is indeed torsion-free.

Indeed, if $xT$ is torsion in $G/T$, then $eT=(xT)^m=x^mT$ for some $m>0$. This means $x^m\in T$, so $(x^m)^n=e$ for some $n>0$. Therefore $x^{mn}=e$ and $x\in T$, so $xT=eT$ and there is no nontrivial torsion element in $T$.

If $G$ is already torsion, then $T=G$ and $G/T$ is the trivial group, which is torsion-free because it has no nontrivial torsion element (having no element different from the identity).

The trivial group $\{e\}$ is indeed both torsion and torsion-free. There's no contradiction.

The proof for the second statement is wrong. you proved that every element of finite order in $G/T$ comes from an element of finite order.

I believe this is a counter example let $G=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ and let $T=\mathbb{Z}/2\mathbb{Z}\times\{0\}$. Clearly $T$ is a normal torsion subgroup of $G$. But $G/T\cong \mathbb{Z}/2\mathbb{Z}$ which is not torsion-free.

Note that there is a difference between a torsion group and the torsion subgroup.

Definition (Torsion group): Let $G$ be a group, we say that $G$ is torsion if every element in $G$ is of finite order.

Definition (the Torsion subgroup): Let $G$ be a group, the torsion subgroup $T$ of $G$ is the subgroup of $G$ containing all elements of finite order. (i.e $T$ is the maximal torsion subgroup of $G$).

And I'm pretty sure that what is actually proved in the book is the following

Theorem: Let $G$ be a group and let $T$ be its torsion subgroup. Then $G/T$ is torsion-free.