[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.
[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]
[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]
