$X$ 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 inclusion $X \hookrightarrow A$ induces a homomorphism on fundamental groups $\pi_1(X) \rightarrow \pi_1(A)$. As you said, pick the element $\gamma$ of $\pi_1(X)$ corresponding to $\{|x| = 2\}$ (i.e., a path traversing this circle once counterclockwise). The image of $\gamma$ in $\pi_1(A)$ represents a generator of $\pi_1(A) \cong \Bbb Z$ (why?), so that $\gamma$ must have infinite order in $\pi_1(X)$ (why?)
(As mentioned in the comments above, this all comes down to having previously computed $\pi_1(S^1)$!)
