How can I show that there cannot exist a homotopy from $\mathcal{C_1}$ to $\mathcal{C_2}$? Consider the diagram below with an annulus $\mathcal{A}$ and two circles in the annulus $\mathcal{C}_1$ and $\mathcal{C}_2$. 
In $\mathbb{R^n}$, there clearly exists a homotopy between any two circles. But if we have this scenario here, where we are restricted by the annulus, it doesn't seem like there exists a homotopy from $\mathcal{C}_1$ to $\mathcal{C}_2$. This is just based on my intuition.
But how can I actually prove this?


I think you mean "homotopy" instead of "homeomorphism".
If that's the case, then you can use the fact that "homotopic" is an equivalence relation, in particular it is transitive.
The circle $\mathcal{C}_2$ is obviously homotopic to the trivial circle. So if you prove that $\mathcal{C}_1$ is not homotopic to the trivial circle, you're done.
You can prove that using complex analysis. $\mathcal{C}_1$ in $\mathcal{A}$ is homeomorphic to $\gamma=\{z \in \Bbb C \ | \ |z|=2\}$ in $D=\{z\in \Bbb C \ | \ 1<|z|<3\}$. Now, $f(z)=\frac 1 z$ is holomorphic on $D$. So by Cauchy's theorem, $\int_\lambda f=0$ for every null-homotopic closed path $\lambda$. 
But $\int_\gamma f\ne 0$ (easy to compute), so $\gamma$ (and thus $\mathcal{C}_1$) are not null-homotopic.
