Let $M$ be a smooth $n$-manifold and let $U\subseteq M$ be any open subset. Define an atlas on $U$ $$\mathcal{A}_{U}=\big\{\text{smooth charts}\;(V,\varphi)\;\text{for}\; M\;\text{such that}\;V\subseteq U\big\}.$$
I must prove that $\mathcal{A}_{U}$ is a smooth atlas for $U$, that is
(1) $U=\bigcup{V}$, where $V$ is the domain of charts such that $V\subseteq U.$
This point is ok.
and
(2) It remains to prove that $\mathcal{A}_U$ is a smooth atlas for $U$.
My attempt. Let $\big(V_1,\varphi_1\big)$, $\big(V_2,\varphi_2\big)\in\mathcal{A}_U$, since they are smooth charts for $M$, that is are charts of maximal atlas of $M$, the maps $$\varphi_2\circ\varphi_1^{-1}\colon\varphi_1\big(V_1\cap V_2\big)\to \varphi_2\big(V_1\cap V_2\big)\quad\text{and}\quad \varphi_1\circ\varphi_2^{-1}\colon \varphi_2\big(V_1\cap V_2\big)\to \varphi_1\big(V_1\cap V_2\big)$$ are $C^{\infty}$.
Since $V_1\subseteq U$, $V_1=U\cap V_1$, then $V_1$ is open in $U$, similary $V_2$ is open in $U$, then $\big(V_1,\varphi_1\big)$ and $\big(V_2,\varphi_2\big)$ are charts of $U$, morever $V_1\cap V_2$ is open in $U$, and then they are $C^{\infty}$ compatible.
Question It's correct?
Your proof is correct, but I think it is uncessary to prove (2). If $\mathcal A$ denotes the maximal smooth atlas of $M$, then any subset $\mathcal A' \subset \mathcal A$ is automatically a smooth atlas on $M' = \bigcup_{(V,\varphi) \in \mathcal A'} V$ which is an open subset of $M$.
It is perhaps worth to mention that $\mathcal A_U$ is a maximal smooth atlas of $U$. To see this, consider any smooth atlas $\mathcal B$ on $U$ containing $\mathcal A_U$.
Let $(W,\psi)$ be any chart in $\mathcal B$. It is compatible with all charts in $\mathcal A_U$. Now let $(\varphi,V) \in \mathcal A$. Its restriction $(\varphi' = \varphi \mid_{V \cap U}, V' = V \cap U)$ also belongs to $\mathcal A$, and since $V \cap U \subset U$, it belongs to $\mathcal A_U$. Since $W \cap V' = W \cap V \cap U = W \cap V$, the charts $(W,\psi)$ and $(\varphi,V)$ have the same transition function as the charts $(W,\psi)$ and $(\varphi',V')$. The latter is smooth, which shows that $(W,\psi)$ is compatible with $(\varphi,V)$.
Hence $(\psi,W)$ is compatible with $\mathcal A$ and we conclude $(\psi,W) \in \mathcal A$. But this shows $(\psi,W) \in \mathcal A_U$ because $W \subset U$.
Therefore $\mathcal B = \mathcal A_U$.