$X$ is contained in an annulus containing a circle in the annulus.Show that $\pi_1(X)$ contains a subgroup isomorphic to $\mathbb{Z}$. In $\mathbb{R^2}$ let $C=\{|x|=2\}$ and $A=\{1<|x|<3\}$, let $X$ be a path connected set such that $C \subset X \subset A$. Show that $\pi_1(X)$ contains a subgroup isomorphic to $\mathbb{Z}$.
Well, it is intuitively obvious but how can one argue rigorously? Can I argue like : $C$ itself can be thought of as a path $\gamma$ in $X$ with $\gamma : [0,1] \to X$ is defined by $\gamma(t)=2e^{2\pi it}$. Since $\gamma$ is not nullhomotopic (My question is : does this fact need proof ? If yes, how should I prove that ?) and hence [$\gamma$] generates an infinite cyclic group which is contained in $\pi_1(X)$... $\blacksquare$ 
Isn't it too handwavy ? thank you in advance for your helpful comments/answers

The fact that it is contained in an annulus means in particular that it doesn't contain the origin. Thus there is a well-defined retract $r:X \to C$ given by
$$r(x) = 2\frac{x}{\|x\|}$$
If you haven't seen it before, the notion of retract is that the inclusion $i:C \to X$ and the surjection (retract) $r:X \to C$ are partial inverses in the sense that $r \circ i = \text{id}_C$. 
It is then a general fact that if we have $Y \subset Z$, with $r:Z \to Y$ a retract, then the induced map $i_*:\pi_1(Y) \to \pi_1(Z)$ is an injection. as $r_*$ is a left inverse.
