Constructing a smooth maximal atlas

Question

Firstly consider the following notations and definitions:

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Notation: Let $(M,\mathfrak{A})$ be a smooth manifold. Then $\color{red}{\tau_M(\mathfrak{A})}$ denotes the topology in $M$ induced by the maximal atlas $\mathfrak{A}$.

Definition: Let $(M,\mathfrak{A})$ a smooth manifold. Suppose that $E$ is a set and $\pi :E\to M$ is surjective map. We say that $(U,\varphi )$ is a local trivialization of $\pi$ with rank $r$ if the following propositions are true:

1. $U\in \tau_M(\mathfrak{A})$
2. $\varphi :\pi ^{-1}[U]\to U\times \mathbb{R}^r$ is a homeomorphism with respect to the topology $\{\pi^{-1}[O]\cap \pi ^{-1}[U]:O\in\tau_M(\mathfrak{A})\}$ in $\pi ^{-1}[U] $ and the product topology of $\tau_M(\mathfrak{A})$ with the standard topology of $\mathbb{R}^r$;
3. $\pi =\pi _1\circ \varphi $ in which $\pi _1:U\times \mathbb{R}^r\to U$ is given by $\pi _1(x,y):=x$;

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My question is: How can I prove the proposition below?

Proposition: Let $(M,\mathfrak{A})$ be smooth manifold with dimension $m$. Suppose that the following propositions are true:

1. $\pi :E\to M$ is surjective map in which $E$ is any set;
2. There's a collection $\big\{(U_i,\varphi _i)\big\}_{i\in I}$ of local trivializations of $\pi$ with rank $r$ such that $M=\cup_{i\in I}U_i$ and for all $i,j\in I$ the map $\varphi _i\circ \varphi^{-1}_j:(U_i\cap U_j)\times \mathbb{R}^r\to (U_i\cap U_j)\times \mathbb{R}^r$ is a smooth map with respect to the obvious product manifold.

Then there's is a smooth maximal atlas $\mathfrak{E}$ such that $(E,\mathfrak{C})$ is a smooth manifold with dimension $m+r$ and $\pi :E\to M$ is a smooth map.

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What I did:

I proved the above proposition. However the length of my proof may hide errors that I'm not being able to notice (so that proposition may even be false).

For this reason I would like to know if there is a simpler proof. Also I would like to know if there is a book that contains the proof of the previous proposition.

If that proposition is true, then $(\pi ,E,M)$ is a vector bundle because we can construct a $\mathbb{R}$-vector space structure in $\pi^{-1}[p]$ for all $p\in U_i$ and $i\in I$ such that $\varphi _p:\pi^{-1}[p]\to \mathbb{R}^r$ given by $\varphi_p (x):=\pi _2\circ \varphi_i (x)$ is an isomorphism of vector spaces in which $\pi _2:U\times \mathbb{R}^r\to \mathbb{R}^r$ is given by $\pi _2(x,y):=y$.

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Below is a sketch of my proof:

Let $i\in I$ be any element.

Let $(V_i,\psi _i)$ be $m$-chart of $(M,\mathfrak{A})$ such that $V_i\subseteq U_i$.

Define $\psi _i\times \text{id}_r:V_i\times\mathbb{R}^r\to \psi [V_i]\times \mathbb{R}^r$ by $(\psi _i\times \text{id}_r)(x,y)=(\psi _i(x),y)$. It's easy to show that this map is smooth.

We can show that $\varphi_i |_{\pi ^{-1}[V_i]}:\pi ^{-1}[V_i]\to V_i\times \mathbb{R}^r$ is a homeomorphism.

Define $\sigma _i:\pi ^{-1}[V_i]\to \psi_i [V_i]\times \mathbb{R}^r$ by $\sigma _i(x):=(\psi _i\times \text{id}_r)\circ \varphi _i(x)$.

We can show that $\big\{(\pi ^{-1}[V_i],\sigma _i)\big\}_{i\in I}$ is a smooth atlas of $E$ (see the hypothesis 2 of the "Proposition" above). So there's a smooth maximal atlas $\mathfrak{E}$ such that $(E,\mathfrak{E})$ is a smooth manifold with dimension $m+r$.

We can also show that $\varphi _i:\pi ^{-1}[U_i]\to U_i\times \mathbb{R}^r$ is a smooth map using the manifold $(E,\mathfrak{E})$.

Since $\pi |_{\pi^{-1}[U_i]}=\pi _1\circ \varphi _i$ for all $i\in I$ and $\pi_1$ is smooth, then we can prove that $\pi :E\to M$ is indeed smooth.