I'm trying to prove that every finite group is generated by its cyclic subgroups.
Let $|G|=n$ and $G=\langle g_1,\ldots,g_n\rangle$. Then $G=\langle g_1,\ldots,g_n\rangle\leq \bigl\langle\langle g_1\rangle,\ldots,\langle g_n\rangle\bigr\rangle\leq G$, since every $\langle g_i\rangle \leq G$. Hence $G=\bigl\langle\langle g_1\rangle ,\ldots,\langle g_n\rangle\bigr\rangle$.
That's the right way to solve this problem? if it's not, how can I show it?
Yes, your reasoning is correct and it generalizes to the case where $G$ is not necessarily finite:
$$G = \langle g : g \in G \rangle \subseteq \langle \langle g \rangle : g \in G \rangle \subseteq G.$$