If $p:E\rightarrow B$ is a fibration (Hurewicz), and we take $b,b'\in B$ in the same path component, I want to prove that the fibers are of the same homotopy type $F_b\simeq F_{b'}$.
I know I should try to find a homotopy equivalence $\alpha:F_b \rightarrow F_{b'}$ and its inverse $\beta:F_{b'}\rightarrow F_b$.
I don't know if it is useful, but what I have thought unitl now is that since there is a path $\gamma:I \rightarrow B$ with $\gamma(0)=b$ and $\gamma(1)=b'$, then for any $a\in F_b$ we may use the homotopy lifting property to obatin a path $f_a:I \rightarrow E$ such that $f_a(0)=a$ and $pf_a = \gamma$, so $f_a(1)\in F_{b'}$.
Of course, with an analogous argument, we can construct paths $g_{a'}:I \rightarrow E$, for any $a'\in F_{b'}$ so that $g_{a'}(0)=a'$, $pg_{a'}=\gamma$ and $g_{a'}(1)\in F_{b}$.
Then I can define the candidates for homotopy equivalences \begin{align} \alpha:F_b &\longrightarrow F_{b'}\\ a &\longmapsto f_a(1) \end{align} and \begin{align}\beta:F_{b'} &\longrightarrow F_b\\ a' &\longmapsto g_{a'}(1) \end{align}
I am finding it difficult to justify why $\alpha$ and $\beta$ are continuous, and to show that $\beta\alpha\simeq id_{F_b}$ and $\alpha\beta\simeq id_{F_{b'}}$.
I don't know if my idea is correct, if there is a simpler way please let me know.