Covering space of an interval is trivializable. We say a covering space $(E,p)\to X$ is trivializable, if there is a space $F$ (with the discrete topology), and a homeomorphism between $\varphi:E\to X\times F$ such that $p=pr_1\circ \varphi$, where $pr_1:X\times F\to X$ is the projection.
Suppose the following proposition holds:

If $B=U\cup V$, both $U$ and $V$ are connected open subsets, and $U\cap V$ is connected. Let $(E,p)$ be a covering space of $B$ such that both $p^{-1}(U)\to U,p^{-1}(V)\to V$ are trivializable, then $(E,p)$ is trivializable.

Show that, if $B\subset\mathbb R$ is an interval, then any covering space of $B$ is trivializable.
The attempt:

We first assert that, if $(E,p)$ is a covering space of $B$, and $B=B_1\cup B_2$, both $B_1,B_2$ are closed, $B_1\cap B_2=\{b\}$, and both $p^{-1}(B_i)\to B_i$ are trivializable, then $(E,p)$ is trivializable.


Let $\varphi_i:p^{-1}(B_i)\to B_i\times F_i$ be the homeomorphisms. Since $B_1\cap B_2=\{b\}$, there is a bijection $\psi:F_2\to F_1$. We define a map
$$f:B\times F_2\to E,\;(u,x)\mapsto\begin{cases}
\varphi_1^{-1}(u,\psi(x)),&\text{if}\; u\in B_1;\\
\varphi_2^{-1}(u,x),&\text{if}\; u\in B_2.
\end{cases}$$
It is not difficult to see $f$ is bijective. We have $f$ is continuous by the pasting lemma and $f^{-1}$ is continuous since $f$ is open. Therefore, $(E,p)$ is trivializable.


For the case that $B$ is a closed and bounded interval $[a,b]$. $\forall x\in B$, there is a neighborhood $U\ni x$ such that $p^{-1}(U)\to U$ is trivializable, then $B$ can be covered by open subsets having this property.
By the lebesgue number lemma, we can choose $a=t_0<t_1<\cdots<t_n=b$ such that $p^{-1}([t_i,t_{i+1}])\to [t_i,t_{i+1}]$ is trivializable, for $i=0,1,\cdots,n-1$. Then we have $p:E\to B$ is trivializable by the assertion above.

But how to show it for other cases?
 A: Let us first show that it is true for $B = [0, \infty)$.
Define $I_n = [n,n+1]$. You know that each $p_n : p^{-1}(I_n)\to I_n$ has a trivializing homeomorphism $\phi_n : p^{-1}(I_n) \to I_n \times F_n$. This gives homeomorphims $h_n : F_n \to F_{n+1}$ determined by $\phi_{n+1}((\phi_n)^{-1}(x,n)) = (h_n(x),n+1)$ and trivializing homeomorphisms
$$\psi_n = h_1^{-1}\dots h_{n-1}^{-1}\phi_n : p^{-1}(I_n) \to I_n \times F_1 .$$
By definition $\psi_n$ and $\psi_{n+1}$ agree on the fiber $p^{-1}(n) = p^{-1}(I_n) \cap p^{-1}(I_{n+1})$. Hence
$$\psi : E \to [0, \infty) \times F_1, \psi(e) = \psi_n(e) \text{ for } e \in p^{-1}(I_n)$$
is a trivializing homeomorphism.
Similarly we can treat $B  = \mathbb R$.
Each non-compact interval $B$ admits a homeomorphism $H : B \to B'$ to one of $B'  = [0, \infty), \mathbb R$. Clearly $p' = Hp : E \to B'$ is a covering. We know that there exists a trivializing homeomorphism $\phi' : E \to B' \times F$. But then
$$\phi = H^{-1}\phi' : E \to B \times F$$
is also a trivializing homeomorphism.
Remark:
A much more general result can be found in Are contractible open subsets always evenly covered?
