Intersections of all subgroups is a nontrivial subgroup, so every element has finite order.

If there is an element $g$ of infinite order, consider the intersection of all subgroups of $(g)$. What is it?

For otherwise there is an infinite cyclic subgroup $(a)$ of $G.$

Due to the non-triviality of the said subgroups, $a^k$ is in the said intersection (for some nonzero integer $k$).

However $(a^{2k})$ is a subgroup of $G$ without having $a^k$ in it !

Suppose there exist an element $g \in G$ of infinite order.

The infinite cyclic groups $(g^n)$ generated by $g^n$ for all $n \in \mathbb N$ are subgroups of $G$.

Any common element of the subgroups $(g^n)$ must have order a multiple of every $n \in \mathbb N$.

$0 \in \mathbb N$ is the only multiple of every $n\in \mathbb N$. (To see this, $0 = n \cdot 0$ for all $n \in \mathbb N$ and any $m \ne 0 \in \mathbb N$ is not a multiple of $m+1 \in \mathbb N$.)

Therefore $g^0 = 1$ is the only element in the intersection of the subgroups $(g^n)$ hence the only element in the intersection of all subgroups of $G$.