[proofplan] We define the proposed maximal atlas by collecting exactly those charts compatible with the original atlas. The main point is to prove that two charts compatible with $\mathcal A$ are compatible with each other; this is done locally by inserting a chart from $\mathcal A$ around each point in the overlap and writing the transition map as a composition of known smooth transition maps. Once this closure under compatibility is established, the collection is a smooth atlas, contains $\mathcal A$, and is maximal by its defining property. [/proofplan]

[step:Define the generated collection of compatible charts] Let \begin{align*} \mathcal A_{\max} := \{(U,\varphi) : (U,\varphi) \text{ is a chart on } M \text{ smoothly compatible with every chart in } \mathcal A\}. \end{align*} We first check that $\mathcal A \subset \mathcal A_{\max}$. If $(U,\varphi) \in \mathcal A$, then $(U,\varphi)$ is smoothly compatible with every chart in $\mathcal A$ because $\mathcal A$ is a smooth atlas. Hence $(U,\varphi) \in \mathcal A_{\max}$. Since $\mathcal A$ covers $M$, the collection $\mathcal A_{\max}$ also covers $M$. [/step]

[step:Prove that any two generated charts are smoothly compatible]Let $(U,\varphi)$ and $(V,\psi)$ be two charts in $\mathcal A_{\max}$. We prove that the transition map \begin{align*} \psi \circ \varphi^{-1}: \varphi(U \cap V) \to \psi(U \cap V) \end{align*} is smooth. The proof for the inverse transition map \begin{align*} \varphi \circ \psi^{-1}: \psi(U \cap V) \to \varphi(U \cap V) \end{align*} is the same with the roles of $(U,\varphi)$ and $(V,\psi)$ exchanged. Fix a point $p \in U \cap V$. Since $\mathcal A$ covers $M$, choose a chart $(W,\theta) \in \mathcal A$ such that $p \in W$. The set $U \cap V \cap W$ is open in $M$, so $\varphi(U \cap V \cap W)$ is an open neighbourhood of $\varphi(p)$ inside $\varphi(U \cap V)$. On $\varphi(U \cap V \cap W)$, we have the identity \begin{align*} \psi \circ \varphi^{-1} = (\psi \circ \theta^{-1}) \circ (\theta \circ \varphi^{-1}). \end{align*} Because $(U,\varphi) \in \mathcal A_{\max}$ and $(W,\theta) \in \mathcal A$, the transition map \begin{align*} \theta \circ \varphi^{-1}: \varphi(U \cap W) \to \theta(U \cap W) \end{align*} is smooth. Because $(V,\psi) \in \mathcal A_{\max}$ and $(W,\theta) \in \mathcal A$, the transition map \begin{align*} \psi \circ \theta^{-1}: \theta(V \cap W) \to \psi(V \cap W) \end{align*} is smooth. Therefore their composition is smooth on $\varphi(U \cap V \cap W)$. Since every point of $\varphi(U \cap V)$ has such an open neighbourhood on which $\psi \circ \varphi^{-1}$ is smooth, the transition map $\psi \circ \varphi^{-1}$ is smooth on all of $\varphi(U \cap V)$. Exchanging the two charts gives smoothness of $\varphi \circ \psi^{-1}$. Thus $(U,\varphi)$ and $(V,\psi)$ are smoothly compatible.[/step]

[guided]We need to prove something that is not built directly into the definition of $\mathcal A_{\max}$: two charts in $\mathcal A_{\max}$ are known to be compatible with every chart in $\mathcal A$, but they are not a priori known to be compatible with each other. Let $(U,\varphi)$ and $(V,\psi)$ be charts in $\mathcal A_{\max}$. Smooth compatibility requires both transition maps to be smooth: \begin{align*} \psi \circ \varphi^{-1}: \varphi(U \cap V) \to \psi(U \cap V), \qquad \varphi \circ \psi^{-1}: \psi(U \cap V) \to \varphi(U \cap V). \end{align*} We prove smoothness of the first map. Fix $p \in U \cap V$. Since $\mathcal A$ is an atlas, it covers $M$, so there exists a chart $(W,\theta) \in \mathcal A$ with $p \in W$. The purpose of this chart is to bridge between $(U,\varphi)$ and $(V,\psi)$ through a chart whose compatibility with both is known. The set $U \cap V \cap W$ is open in $M$, and $\varphi: U \to \varphi(U)$ is a homeomorphism. Hence $\varphi(U \cap V \cap W)$ is an open neighbourhood of $\varphi(p)$ in $\varphi(U \cap V)$. On this neighbourhood, the transition map factors as \begin{align*} \psi \circ \varphi^{-1} = (\psi \circ \theta^{-1}) \circ (\theta \circ \varphi^{-1}). \end{align*} Now each factor is smooth for a reason encoded in the definition of $\mathcal A_{\max}$. Since $(U,\varphi)$ is compatible with every chart in $\mathcal A$, and $(W,\theta) \in \mathcal A$, the map \begin{align*} \theta \circ \varphi^{-1}: \varphi(U \cap W) \to \theta(U \cap W) \end{align*} is smooth. Since $(V,\psi)$ is compatible with every chart in $\mathcal A$, and $(W,\theta) \in \mathcal A$, the map \begin{align*} \psi \circ \theta^{-1}: \theta(V \cap W) \to \psi(V \cap W) \end{align*} is smooth. Therefore their composition is smooth on $\varphi(U \cap V \cap W)$. Because this construction works around every point $p \in U \cap V$, the map $\psi \circ \varphi^{-1}$ is locally smooth at every point of its domain. Smoothness of maps between open subsets of $\mathbb R^n$ is local, so $\psi \circ \varphi^{-1}$ is smooth on $\varphi(U \cap V)$. Repeating the same argument with $(U,\varphi)$ and $(V,\psi)$ interchanged proves that $\varphi \circ \psi^{-1}$ is smooth on $\psi(U \cap V)$. Hence the two charts are smoothly compatible.[/guided]

[step:Conclude that the generated collection is a smooth atlas] The collection $\mathcal A_{\max}$ covers $M$ because it contains $\mathcal A$. By the previous step, every pair of charts in $\mathcal A_{\max}$ is smoothly compatible. Therefore $\mathcal A_{\max}$ is a smooth atlas on $M$. [/step]

[step:Verify maximality of the generated atlas] Let $(Z,\zeta)$ be a chart on $M$ smoothly compatible with every chart in $\mathcal A_{\max}$. Since $\mathcal A \subset \mathcal A_{\max}$, the chart $(Z,\zeta)$ is smoothly compatible with every chart in $\mathcal A$. By the definition of $\mathcal A_{\max}$, this implies $(Z,\zeta) \in \mathcal A_{\max}$. Thus no chart can be added to $\mathcal A_{\max}$ while preserving smooth compatibility with all charts already in $\mathcal A_{\max}$. Hence $\mathcal A_{\max}$ is a maximal smooth atlas containing $\mathcal A$. [/step]