Smoothness of a manifold implies all maps of its altas' charts are diffeomorphism? I need to show that :

On any smooth manifold $(M,A)$ all chart maps are $C^{\infty}$-diffeomorphisms.

Definitions : Let $M$ be a Hausdorff second countable topological space. Then  a pair $(U, \phi)$ is called a chart for $M$ if $U \in M$ is open and $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ be a homeomorphism. The set $\{\ (U_{\alpha}, \phi_{\alpha}) \}$ is called a topological atlas for $M$ if $\cup U_{\alpha} = M$ and then $(M,A)$ will be called a topological manifold. For $(M,A)$, two charts $(U, \phi), (V, \psi) \in A$ are called $C^{\infty}$-related if $U \cap V = \emptyset$ or if $U \cap V \ne \emptyset$ then $\psi \circ \phi^{-1} : \phi(U \cap V) \subset \mathbb{R}^n \to \psi(U \cap V) \subset \mathbb{R}^n$ be $C^{\infty}$-diffeomorphism. If all charts in $(M,A)$ are pairwise $C^{\infty}$-related then $(M,A)$ is called a smooth manifold.
My attempt based on the definitions above to answer the question in the first paragraph : , The question asks that if for any $(U, \phi) \in A$, $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ is a $C^{\infty}$-diffeomorphism. So because $\phi \circ \phi^{-1} = \operatorname{id}$ is $C^{\infty}$-diffeomorphism so is $\phi$.
Am I in a right track and if so how to make this rigorous?
Added : Can I go with this approach that I add $\operatorname{id}$ to atlas and say that $\phi = \phi \circ id$ so $\phi$ is diffeomorphism?
 A: Consider $\mathcal{A}_U = \{(U,\phi)\}$. It is a smooth atlas on $U$ (show it), and therefore, $(U,\mathcal{A}_U)$ is a smooth manifold.
Similarly,  $\mathcal{A}_{\phi(U)} = \{\phi(U),\mathrm{id}_{\phi(U)}\}$, is a smooth atlas on $\phi(U)$ and it follows that $(\phi(U),\mathcal{A}_{\phi(U)})$ is a smooth manifold.
Let us look at the homeomorphism $\phi \colon U \to \phi(U)$ through the only charts of our atlases. It is given by
$$
\mathrm{id}_{\phi(U)}\circ \phi \circ (\phi)^{-1} = \mathrm{id}_{\phi(U)}
$$
and is therefore smooth. Similarly, the inverse map $\phi^{-1} \colon \phi(U) \to U$ reads, in the charts, as follows
$$
\phi \circ \phi^{-1} \circ (\mathrm{id}_{\phi(U)})^{-1} = \mathrm{id}_{\phi(U)}
$$
which is again smooth. It follows that $\phi\colon U \to \phi(U)$ is a smooth homeomorphism with smooth inverse between two smooth manifolds, and is hence a diffeomorphism.
Edit
Smoothness is a notion defined for functions from some open subset of $\Bbb R^p$ with range into some other open subset of $\Bbb R^q$. In order to extend this definition for maps between manifolds, one has to find some trick to go back to the Euclidean setting. The trick is the following.
Let $f\colon M \to N$ be a map between smooth manifolds. If $(U,\phi)$ is a chart of $M$, $(V,\psi)$ a chart of $N$, and if $f(U) \subset V$, then the map
$$
\psi \circ f \circ \phi^{-1} \colon \phi(U) \to \psi(V)
$$
goes from an open subset of an Euclidean space onto another one. Here, we have a notion of smoothness.

Definition. A map $f\colon M\to N$ is said to be smooth if for any charts $(U,\phi)$ and $(V,\psi)$ as above, the map $\psi\circ f \circ \phi^{-1} \colon \phi(U) \to \psi(V)$ is smooth.

A: Building on the answer by Lee Mosher, you have the following definition:
A map $f$ from an open subset $U\subset M$ to $\mathbb{R}^k$ is smooth if and only if its composition with each chart inverse $\phi^{-1}$ associated to $W$ with $U\cap W \neq \emptyset$ is smooth as map between Euclidean spaces. If our manifold is $n$-dimensional, this is saying to look at
$$ f\circ\phi^{-1}: \phi(U\cap W)\subset\mathbb{R}^n\to \mathbb{R}^k. $$
Since your map $f$ is a chart map, what can you say about the expression above? Which definition out of the ones you wrote down does it look like?
A: Let us go step by step.
We need to establish (recall) some preliminary results.
Step 1. It is easy to show that if $M$ is a smooth manifold with atlas $\{\ (U_{\alpha}, \phi_{\alpha}) \}$, and $W$ is an open set in $M$, then $W$ is a smooth manifold with atlas $\{\ (W\cap U_{\alpha}, \phi_{\alpha}|_{W\cap U_{\alpha}}) \}$
Step 2. Let $M$ be smooth manifold with atlas  $\{\ (U_{\alpha}, \phi_{\alpha}) \}$, $N$ be smooth manifold with atlas $\{\ (V_{\beta}, \psi_{\beta}) \}$. Then

A function $f: M \to N$ is $C^{\infty}$ if $f$ is continuous and, for all $\alpha$ and $\beta$, the function $\psi_\beta \circ f \circ \phi_\alpha^{-1}: \phi_\alpha(U_\alpha \cap f^{-1}(V_\beta)) \to \psi_\beta({V_\beta})$ is $C^{\infty}$.

Note that $\phi_\alpha(U_\alpha \cap f^{-1}(V_\beta))$ is an open set in $\Bbb R^n$  and $\psi_\beta({V_\beta})$ is an open set in $\Bbb R^k$ (for some $k$, dimension of $N$), so it is clear what means $\psi_\beta \circ f \circ \phi_\alpha^{-1}: \phi_\alpha(U_\alpha \cap f^{-1}(V_\beta)) \to \psi_\beta({V_\beta})$ to be $C^{\infty}$.
Note also that, if $U_\alpha \cap f^{-1}(V_\beta) = \emptyset$, then
$\phi_\alpha(U_\alpha \cap f^{-1}(V_\beta))= \emptyset$ and $\psi_\beta \circ f \circ \phi_\alpha^{-1}: \phi_\alpha(U_\alpha \cap f^{-1}(V_\beta)) \to \psi_\beta({V_\beta})$ is trivially $C^{\infty}$.
Step 3. The space $\Bbb R^n$ is smooth manifold with the standard (trivial) atlas $\{(\Bbb R^n, id_{\Bbb R^n} )\}$. So, by Step 1, if $W$ is an open set in $\Bbb R^n$, $W$ is a smooth manifold with atlas $\{(W, id_{W} )\}$.
So, in Step 2, if $V$ is an open set of $\Bbb R^k$ and $N= V$, we can simplify the definition of $f$ being $C^{\infty}$.
In fact, let $M$ be smooth manifold with atlas  $\{\ (U_{\alpha}, \phi_{\alpha}) \}$, then

A function $f: M \to V$ is $C^{\infty}$ if $f$ is continuous and, for all $\alpha$, $ f \circ \phi_\alpha^{-1}: \phi_\alpha({U_\alpha} \cap f^{-1}(V)) \to V $ is $C^{\infty}$.

and

A function $g: V \to M$ is $C^{\infty}$ if $g$ is continuous and, for all $\alpha$, $\phi_\alpha \circ g : V \cap g^{-1}(U_\alpha) \to \phi_\alpha(U_\alpha)$ is $C^{\infty}$.

Step 4. We want to prove that,

for each  $(U, \phi) \in A$, $\phi : U \subset M \to \phi(U) \subset \mathbb{R}^n$ is a $C^{\infty}$-diffeomorphism.

It means $\phi$ is bijective and $\phi$ and $\phi^{-1}$ are $C^{\infty}$-functions. The proof that $phi$ is trivial from the fact that $(U, \phi)$ is a chart.
Now, to prove that   $\phi : U \to \phi(U) \subset \mathbb{R}^n$ is a $C^{\infty}$, we must consider U as smooth manifold (by Step1) and, since $\phi(U)$ is an open set in $\Bbb R^n$ and $\phi^{-1}(\phi(U))=U$, then by Step 3, it is enough to prove that , for all $\alpha$,  $\phi \circ \phi_\alpha^{-1} : \phi_\alpha(U_\alpha \cap U) \to \phi(U) \subset \mathbb{R}^n$ is $C^{\infty}$. If $U_\alpha \cap U =\emptyset$, it is trivially true that  $\phi \circ \phi_\alpha^{-1}$ is $C^{\infty}$. If $U_\alpha \cap U \neq \emptyset$, it is a trivial consequence of the charts $(U,\phi)$ and $(U_\alpha,\phi_\alpha)$ being $C^{\infty}$-related.
Now, to prove that   $\phi^{-1} :  \phi(U) \subset \mathbb{R}^n \to U$ is a $C^{\infty}$, is similar. We must consider U as smooth manifold (by Step1) and, since $\phi(U)$ is an open set in $\Bbb R^n$ and, for any $\alpha$,
$(\phi^{-1})^{-1}(U_\alpha)=(\phi^{-1})^{-1}(U \cap U_\alpha)=\phi(U \cap U_\alpha)$,
by Step 3, it is enough to prove that , for all $\alpha$,  $\phi_\alpha\circ \phi^{-1} : \phi(U \cap U_\alpha) \to \phi_\alpha(U_\alpha)$ is $C^{\infty}$. If $U\cap U_\alpha =\emptyset$, it is trivially true that  $\phi_\alpha \circ \phi^{-1}$ is $C^{\infty}$. If $U\cap U_\alpha \neq \emptyset$, it is a trivial consequence of the charts $(U_\alpha,\phi_\alpha)$ and $(U,\phi)$ are $C^{\infty}$-related.
