Let $X$ be a nice enough space, say paracompact. Let $X^{*}$ be the closed cylinder with respect to $X$. Assume $p:E\rightarrow X^{*}$ is fiber bundle.
Since the second factor of the cylinder is compact, we can as a result find can find open cover $(V_i)_{i=1}^{\infty}$ be an open cover for $X$ such that $E|_{V_i\times I}$ is trivial. Paracompactness of $X$ suggests that we can find $(\varphi_i)$ partitions of unity subbordinate to $V_i$.
So their sum is $\equiv 1$ and support is contained $V_i$ and image is contained in $[0,1]$. Set $\psi_i=\varphi_1+...+\varphi_i$. Notice $\sum_i\psi_i \equiv 1$.
Assume that $E$ is trivial over $V_i\times I$. Let $X_i$ be the graph of $\psi_i$. Then as stated in page 21 in https://pi.math.cornell.edu/%7Ehatcher/VBKT/VB.pdf, the natural projection homeomorphism $X_i\rightarrow X_{i-1}$ that takes $(x,\psi_{i}(x))$ to $(x,\psi_{i-1}(x))$ lifts to a homeomorphism $p^{-1}(X_i)\rightarrow p^{-1}(X_{i-1})$.
I am having a hard time checking that the map $p^{-1}(X_i)\rightarrow p^{-1}(X_{i-1})$ is well defined and a homeomorphism. Could someone please help?
Is the support of $\psi_i$ contained $U_i$? I don't think so but I don't see otherwise how to lift the map. I think its contained in the union up to $U_0,...,U_{i-1}$.
It seems to me that we can lift to a homeomorphism provided $X_i$ is the graph of $\psi_i$ restricted to $U_i$.
I am not sure how the map is even constructed. I tried 'pulling back' by the trivialization but I wasn't able to show continuity which I tried by showing local continuity.
The first thing to note is that $\psi_{i-1} = \psi_i-\varphi_i$ and the map $$\xi_i\colon X_i\cap (V_i\times I)\to V_i\times I, (x,\psi_i(x))\mapsto (x,\psi_{i-1}(x)) = (x,\psi_i(x)-\varphi(x))$$ is continuous.
Now suppose the fiber is $F$ and let $\eta_i\colon p^{-1}(V_i\times I)\xrightarrow{\cong} V_i\times I\times F$ be a trivialization of $E$ over $V_i\times I$. Suppose that $\eta_i(e) = (x,\psi_i(x),f)$. If $x\not\in V_i$, then $(x,\psi_{i}(x)) = (x,\psi_{i-1}(x))$, since the support of $\varphi_i$ is contained in $V_i$. In other words, if $e\in p^{-1}(X_i)\setminus p^{-1}(V_i\times I)$, then $e\in p^{-1}(X_{i-1})$. Thus, if we define $$q_i(e) = \begin{cases}e&\text{if }e\not\in p^{-1}(V_i\times I),\\ (\eta_{i})^{-1}(x,\psi_{i-1}(x),f)= (\eta_{i})^{-1}(x,\psi_{i}(x)-\varphi_i(x),f) &\text{if }e\in p^{-1}(V_i\times I), \end{cases}$$ this truly defines a map $q_i\colon p^{-1}(X_i)\to p^{-1}(X_{i-1})$ (clearly $(\eta_{i})^{-1}(x,\psi_{i-1}(x),f)\in p^{-1}(X_{i-1})$ by construction).
To check continuity first note that by the above $q_i$ is continuous on $p^{-1}(X_i)\cap p^{-1}(V_i\times I)$, since there it is given by $$(\eta_{i})^{-1}(x,\psi_{i-1}(x),f) = (\eta_{i})^{-1}(\xi_i(x,\psi_{i}(x)),f),$$ which is a composition (and product) of continuous maps. Then $q_i$ is also continuous on $p^{-1}(X_i)\cap p^{-1}(\mathrm{supp}(\varphi_i)\times I)$, which is closed in $p^{-1}(X_i)$. On the other hand, if $Z_i = \{x\in X:\varphi_i(x)=0\}$, then $q_i$ is the identity/inclusion on $p^{-1}(X_i)\cap p^{-1}(Z_i\times I)$, hence continuous. Since also $p^{-1}(X_i)\cap p^{-1}(Z_i\times I)$ is closed in $p^{-1}(X_i)$ and $$p^{-1}(X_i) = (p^{-1}(X_i)\cap p^{-1}(Z_i\times I))\cup (p^{-1}(X_i)\cap p^{-1}(\mathrm{supp}(\varphi_i)\times I))$$ we see that $q_i$ is continuous on the whole of $p^{-1}(X_i)$.