Theorem:
Every infinite group $G$ has a subgroup $H$ that is non-trivial ($H \ne G, \lbrace e \rbrace$).
Proof: This will be a proof by contradiction. So we will assume every subgroup is trivial and bring the cyclic subgroups to the table.
For the non-identity $\forall x \in G$, $\langle x \rangle \ne \lbrace e \rbrace$. Hence $\langle x \rangle = G.$ Thus every non-identity element of $G$ must be a generator of $G$. Then it must be possible to write any element as the exponent of another. For $\forall y \in G$,
$$y =x^n.$$
Since $x^2 \in G$ by closure, it is also a generator and we must be able to write $x$ as an exponent of $x^2$. Yet this is not possible unless our group is finite. So our assumption must be false and our conjecture must be true. $\square$
Is my proof watertight? And is it rigorous enough? Thanks for your reviews.
Best Answer
The idea is right, but you should avoid claims like
since this doesn't give your reason for the "unless."
In your case, you perhaps could explicitly note that $G = \langle x \rangle$ implies $x$ cannot have finite order. Then show that $x \in \langle x^2\rangle$ implies $x$ has finite order.