Every numerable principal $G$-bundle is regular over $SB$ Studying Fibre Bundles from Husemoller, I stumble across the following problem in chapter $8$.
Definition: An open covering $\lbrace U_i \rbrace_{i \in S}$ of a topological space is numerable provided there exists a locally finite partition of unity $\lbrace u_i \rbrace_{i \in S}$ such that $\overline{u_i^{-1}(0,1]} \subset U_i$ for each $i \in  S$.
Definition: A principal $G-$bundle $\xi$ over a space $B$ is numerable provided there is a numerable $\lbrace U_i \rbrace_{i \in S}$ of $B$ such that $\xi|U_i$ is trivial for each $i \in S$.
Definition: A bundle $\xi$ over a suspension $SB = \frac{B \times [-1,1]}{B \times \lbrace -1 \rbrace \cup B \times \lbrace 1 \rbrace}$ us regular provided, for dome $\varepsilon > 0$, $\xi|B(-\varepsilon, 1]$ and $\xi|B[-1,\varepsilon)$ are trivial.
In the sentence following the latter definition the author states without explaining:

Every numerable principal $G$-bundle is regular over $SB$.

Is this a straightfoward consequence? Requiring numerable means having an open covering and trivial restriction. Is there an easy way to glue them all and find $\varepsilon$ or another approach is needed?
Any help or reference would be appreciated.
 A: You didn't say in your question what the notations $B[-1,\varepsilon)$ and $B(-\varepsilon,1]$ mean, but as you noted in a comment, they stand for the images of $B\times [-1,\varepsilon)$ and $B\times (-\varepsilon,1]$ under the quotient map, respectively.
For any $\varepsilon\in (0,1)$, both of these sets are contractible: For example, the straight-line homotopy $H\colon B\times (-\varepsilon,1]\times [0,1]\to B\times (-\varepsilon,1]$ defined by $H((b,s), t) = (b,s+t(1-s))$ descends to the quotient to give a homotopy $\widetilde H\colon B(-\varepsilon,1]\times [0,1]\to B(-\varepsilon,1]$ satisfying $\widetilde H((\cdot,\cdot),0) = \text{Id}$ and $\widetilde H((\cdot,\cdot),1) = \text{constant}$.
As @Tyrone mentioned in a comment, every numerable bundle over a contractible space is trivial. (Corollary 10.3 in Husemoller's 3rd edition shows that every numerable principal bundle is trivial, and it follows that every fiber bundle is trivial by applying the corollary to the associated principal bundle.) Thus every numerable bundle over a suspension satisfies the regularity condition for each $\varepsilon\in (0,1)$.
