Example of a specific type of infinite group Give example (if exists) of an infinite non-cyclic group  such that every non-trivial subgroup  of it has  finite index ; off-course the group must be non-abelian ; motivated form $G/H$ is a finite group so $G\cong\mathbb Z$ 
 A: I think there is no such group.  First, $G$ is a torsion-free and virtually cyclic.  For, if $g$ is a non-trivial element of $G$, then $\langle g\rangle$ has finite index in $G$, so $G$ is virtually cyclic.  And, since $G$ is infinite, the element $g$ must have infinite order. As $g$ was arbitrary, $G$ is torsion-free.  Also, note that $G$ is finitely generated. (See below.)
Now, if $G = \langle x_1,\ldots, x_n\rangle$, then $C_G(x_i)$ is non-trivial, so it has finite index  in $G$.  Therefore, $Z(G) = \bigcap_{i=1}^n C_G(x_i)$ has finite index in $G$, so the derived subgroup $[G,G]$ of $G$ is finite, by Schur's Theorem.  Since $G$ is torsion-free, it follows that $G$ is abelian.  Now $G$ must be infinite cyclic.
ADDED
Here is the requested proof that $G$ is finitely generated.  We have that $\langle g\rangle$ is a subgroup of finite index in $G$, where $g$ is any non-trivial element of $G$. Therefore, we can write $G$ as the disjoint union of cosets:
$$G = y_1\langle g\rangle\cup y_2\langle g\rangle\cup\cdots\cup y_m\langle g\rangle,$$
for suitable elements $y_1, y_2,\ldots, y_m\in G$, where $m$ is the index of $\langle g\rangle$ in $G$. (We could take, for example, $y_1 = 1\in\langle g\rangle$, but it's not important here.)  This means that an arbitrary element $u$ in $G$ belongs to (exactly) one of those cosets, say, $u\in y_i\langle g\rangle$.  Therefore, $u = y_ig^s$, for some integer $s$.  This shows that every element of $G$ can be written as a product of powers of elements of the set $\{ y_1, y_2,\ldots, y_m, g\}$, so $G$ is generated by this finite set.
(Exercise. Show that a group is finitely generated if it has a finitely generated subgroup of finite index.)
