This is the lemma 7.1 of Metric structures in differential geometry of Walschap.
Lemma 7.1 : Let $\xi$ denote a vector bundle over an $n$-dimensional manifold $B$. Then $B$ can be covered by $n+1$ sets $U_o,\dots, U_n$ , where each restriction $\xi|_{U_i}$ is trivial.
proof : Choose an open cover of $B$ such that $\xi$ is trivial over each element. It is a well-known theorem in topology that this (and in fact any) cover of an $n$-dimensional manifold $B$ admits a refinement $\{ A_\alpha \}_{\alpha \in A}$ with the property that any point in $B$ belongs to at most $n + 1$ $V_\alpha$'s. Let $\{ \phi_\alpha \}$ be a partition of unity subordinate to this cover, and denote by $A_i$ the collection of all subsets of $A$ with $i + 1$ elements. Given $a = \{ \alpha_0, \dots, \alpha_i \} \in A_i$, denote $W_a$ the set consisting of those $b\in B$ such that $\phi_\alpha(b) < \phi_{\alpha_0}(b), \dots, \phi_{\alpha_i}(b)$ for all $\alpha \neq \alpha_0, \dots, \alpha_i$.
To clarify, we may rewrite $W_a$ as $$ W_a = \{ b \in B \mid \forall \alpha \in a, \forall \beta \in A \setminus a, \phi_\beta(b) < \phi_\alpha(b) \}. $$
Then the author conclude that $W_a$ is open for all $a$ but I don't understand why. When I try to write the definition of $W_a$ with union and intersection of simpler opens I arrive to a infinite intersection which is not obviously open.
The following proof is incomplete, containing an extra unwaranted assumption. But maybe someone can fill in the gap.
For ease of notation, I'll use $\phi$ to denote the minimum of $\phi_{\alpha_1},... \phi_{\alpha_i}$. So, $b\in W_{\alpha}$ iff $\phi_{\alpha}(b) < \phi(b)$ for all $\alpha\neq \alpha_1,...,\alpha_i$.
Suppose $b\in W_\alpha$. We need to find an open neighborhood $U$ of $b$ with $U\subseteq W_\alpha$. To that end, let $V_{b_1},....,V_{b_{n+1}}$ denote the $n+1$ elements of $\{V_\alpha\}_{\alpha \in A}$ which contain $b$.
Extra assumption: The sets $V_1,...,V_{n+1}$ are distinct.
Set $V = V_{b_1}\cap ... \cap V_{b_{n+1}}$. Then $V$ is an open set, and $b\in V$, so it's non-empty. Note that every element of $V$ is contained in $n+1$ of the $V_\alpha$ (namely, $V_{b_1},..., V_{b_{n+1}}$, so these are all the sets in $\{V_{\alpha}\}$ which contain any point in $V$.
Since $b\in W_\alpha$, $\phi_{b_j}(b) < \phi(b)$ for all $j = 1,...,n+1$. For each $j=1,....,n+1$, let $U_j\subseteq V_{b_j}$ be the open set $\{c\in V_{b_j}: \phi_{b_j}(c) < \phi(c)\}$.
Consider the set $U = U_1\cap...\cap U_{n+1}\subseteq V$. This is an intersection of finitely many open sets, so is open. Also note that $b\in U_j$ for all $j$, so $U$ is a neighborhood of $b$. Moreover, since $U\subseteq V$, any element $c\in U$ is in $V_{b_1},...,V_{b_{n+1}}$, but is in no other $V_\alpha$.
We claim that $U\subseteq W_\alpha$. To see this, we must show for any $c\in U$ and for any $\alpha\notin \{\alpha_1,...,\alpha_i\}$, that $\phi_{\alpha}(c) < \phi(b)$. We break into cases depending on whether $\alpha \in \{b_1,....,b_{n+1}\}$ or not. If $\alpha$ is in this set, then by definition of $U_j$, we know that $0\leq \phi_{b_j}(c) < \phi(c)$. If $\alpha$ is not in this set, then $c\notin V_\alpha$, so $\phi_{\alpha}(c) = 0 < \phi(c)$. This completes the proof.