Given a continuous map $p:E\rightarrow B$ where $B$ is given by a colimit of $B_{0}\subseteq B_{1}\subseteq B_{2}\subseteq\dots$. We get the canonical induced map
$$ colim_{n\in\mathbb{N}} p^{-1}(B_{n})\xrightarrow{v} E$$
It is a continuous and bijective map. The background discussion is: Is it a homeomorphism? What are sufficient conditions for it to be a homeomorphisms/homotopy equivalence/weak homotopy equivalence?
What I can say is, that it is a weak equivalence, if $p$ is a Serre fibration and $E$ is nonempty. For in this case choose an element of $e$ and let $F:=p^{-1}(\{p(b)\})$. The long exact sequence for Serre fibrations yields
$$\require{AMScd} \begin{CD} \dots @>>> \pi_{n}(F, e) @>>> \pi_{n}(colim_{n\in\mathbb{N}}p^{-1}(B_{n}),e) @>>> \pi_{n}(B, p(e)) @>>> \dots \\ & @VV id V @VV \pi_{n}(v) V @VV id V \\ \dots @>>> \pi_{n}(F,e) @>>> \pi_{n}(E,e) @>>> \pi_{n}(B,p(e)) @>>> \dots \\ \end{CD} $$
Now we can apply the five lemma.
My two questions are:
Did I use a sledgehammer to crack a nut? Is there a more elementary way to see this?
Can this statement be generalised or strengthened?
As long as every compact subspace of $B$ is contained in some $B_n$ (which is automatically the case if $B$ is $T_1$; see Compact subset in colimit of spaces), then $v$ is a weak equivalence. Indeed, note that this implies every compact subspace $K$ of $E$ is contained in some $p^{-1}(B_n)$, since $p(K)$ is compact. Since homotopy groups are defined in terms of maps out of compact spaces, this immediately implies $v$ is a weak equivalence.
Here is another criterion you may find helpful: if the interiors of the $B_n$ cover $B$, then $v$ is a homeomorphism. Indeed, since $p$ is continuous, this implies the interiors of the $p^{-1}(B_n)$ cover $E$, which implies $v$ is a homeomorphism.
It is not true in general that $v$ is a weak equivalence. For instance, let $B=\mathbb{N}$ with the indiscrete topology and $B_n=\{0,\dots,n-1\}$. Let $E=\mathbb{R}$ and let $p:E\to B$ be a map such that $p^{-1}(\{n\})$ is dense in $\mathbb{R}$ for all $n$. Then $p^{-1}(B_n)$ is totally disconnected for all $n$ (since its complement is dense) and thus the colimit $\operatorname{colim} p^{-1}(B_n)$ is totally path-disconnected (since any path in the colimit would be contained in $p^{-1}(B_n)$ for some $n$ by compactness). In particular, then, $v$ is not an isomorphism on $\pi_0$.