I was thinking of taking two subgroups with $n$ elements and showing that they both are the same subgroup.
I know that if $M$ ($2\times2$ matrix) is any element of $SO(2)$, $M^{-1} = M^t$. And $\det M =1$.
I don't know how to proceed.
Any help would be appreciated.
You can think of $\mathrm{SO}(2)$ as the unit circle under complex multiplication. Therefore, for any point $g_i$ in your group, you get a real number $t_i$ such that $e^{it_i}=g_i$. In other words, consider this map:
$$\varphi: (\mathbb{R},+) \to (\mathbb{C}^{\times},\cdot)\cong\mathrm{SO}(2)$$
where $\mathbb{C}^{\times}$ is the unit circle in the plane. This is an epimorphism of groups, i.e. it is surjective and it is a homomorphism.
Now since your subgroup of $\mathrm{SO}(2)$ is finite, take $g_i$ to be the element with the smallest $|t_i|$ in $\mathbb{R}$. In other words:
$$g_i = \varphi(\min\{t_j\in [0,2\pi): t_j=\varphi^{-1}(\{g_j\}) \})$$ Show that $g_i$ generates your group. The proof is similar to when you want to prove that any subgroup of $\mathbb{Z}$ is generated by a $m \in \mathbb{Z}$.
Once you have shown that your subgroup is generated by $g_i$ and therefore it is cyclic, note that any two cyclic groups of the same order are isomorphic. And that ends the proof.