I like your idea to use Prop 1 and 2!
Part of the power of algebraic topology comes from translating the topology (which is frequently quite difficult to get your hands on, since everything is continuous, up to homotpy, etc.) into algebra (which is frequently quite easy to get your hands on, since everything is discrete).
One example of this is the duality of covering spaces and subgroups of $\pi_1$.
Covering spaces of $X$ (up to isomorphism) are in bijection with subgroups of $\pi_1 X$ (up to conjugacy)!
This happens since if $Y$ covers $X$, then $\pi_1 Y$ is a subgroup of $\pi_1 X$.
Moreover, this correspondence is "direction reversing" in the sense that bigger covering spaces correspond to smaller subgroups (notice this makes sense, since the universal cover corresponds to the trivial subgroup)
Formally, say $X_1$ and $X_2$, covering spaces of $X$, correspond to $H_1$ and $H_2$, subgroups of $\pi_1 X$.
Then $X_1$ covers $X_2$ if and only if $H_1 \leq H_2$
Now, you're given a space $X$ where $\pi_1 X = \mathbb{Z}$. So we have a very good understanding of the subgroups of $\pi_1 X$! That means, using this correspondence, we must have a very good understanding of the covering spaces of $X$!
This is remarkable, since covering spaces are extremely complicated objects, but we can understand them by simply looking at subgroups of $\mathbb{Z}$, which are classified by natural numbers!
So then, with all this in mind, here's a hint for solving your problem:
We know that $X_1$ and $X_2$ cover $X$. So $\pi_1 X_1 = m \mathbb{Z}$ and $\pi_1 X_2 = n \mathbb{Z}$ are subgroups of $\mathbb{Z}$.
We want to find a space $Y$ which covers $X_1$ and $X_2$. But by our (direction reversing!) correspondence, it suffices to construct a subgroup of $\mathbb{Z}$ which is contained in both $m \mathbb{Z}$ and $n \mathbb{Z}$. Asking that $Y$ not be simply connected is asking that this subgroup is nontrivial.
Can you find such a subgroup?
Once you have a subgroup $H$, we can construct the space $Y$ by looking at $\widetilde{X} \big / H$. That is, take the universal cover of $X$, and quotient out by the natural action of $H$ on it. You can find more details about this construction here, for instance.
I hope this helps ^_^
Best Answer
$S^1$ is the quotient of $R$ by $f(x)=x+1$, $Z/2$ acts on $S^1$ by the action induced by the translation $g(x)=x+1/2$ and the quotient of this action is still $S^1$.