Thank you for helping me. I have written down my proof and would be very grateful to anyone who could read it through for me.
Best wishes and thank you!
$\underline{(1) \Rightarrow (2):}$
Put $F=A$ and give it the discrete topology. For each $\alpha \in A$ define
$$h_\alpha:V_\alpha \to U \times A; \; h_\alpha:y \mapsto (p(y),\alpha).$$
For each $\alpha \in A$, the map $h_\alpha$ is continuous.
Indeed, given $\alpha \in A$, the first component map is just $p|_{V_\alpha}$ which is continuous by hypothesis and the second component map is the constant map $y\mapsto \alpha$ which is continuous. Thus $h_\alpha$ continuous.
Now define a map $h:p^{-1}(U) \to U\times A$ by
$$h|_{V_\alpha} := h_\alpha \text{ for each } \alpha \in A.$$
Note that is indeed a map as the $V_\alpha$ are disjoint and $p^{-1}(U)=\bigcup_{\alpha \in A}V_\alpha$.
It is immediate from the definition of $h$ that $p|_{p^{-1}(U)}=\pi_1\circ h$.
It remains to prove that $h$ is a homeomorphism.
$\bullet \; h$ is continuous:
As each $h_\alpha$ is continuous, $h$ is continuous by the "pasting lemma".
$\bullet \; h$ is injective:
Suppose $h(y_1)=h(y_2)$ where $y_1 \in V_{\alpha}$ and $y_2 \in V_{\alpha'}$ for some $\alpha, \alpha' \in A$.
Then $p(y_1)=p(y_2)$ and $\alpha=\alpha'$.
Thus $p|_{V_\alpha}(y_1)=p|_{V_\alpha}(y_2)$ and so $y_1=y_2$ as $p|_{V_\alpha}$ is injective by hypothesis.
$\bullet \; h$ is surjective:
Let $(x,\alpha) \in U \times A$ be given. By hypothesis, $p|_{V_\alpha}$ is surjective so there exists $y \in V_\alpha$ such that
$$h(y)=(p(y),\alpha)=(x,\alpha).$$
$\bullet \; g:=h^{-1}:U \times A \to p^{-1}(U)$ is continuous:
$U \times A$ is the union of disjoint, open sets $\{U \times \{\alpha\} : \alpha \in A\}$.
As such, by the "pasting lemma", it suffices to prove that, given $\alpha \in A$, $ \;g|_{U\times\{\alpha\}}$ is continuous.
Indeed, $g|_{U\times\{\alpha\}} = p|_{V_\alpha}^{-1} \circ \pi_1$ which is continuous by hypothesis.
$\\$
$\underline{(2) \Rightarrow (1):}$
For each $\alpha \in F$ put $V_\alpha=h^{-1}(U\times \{\alpha\})$.
Clearly the $V_\alpha$ are disjoint and $p^{-1}(U)=\bigcup_{\alpha \in F} V_\alpha$ (preimage of a partition).
Moreover, given $\alpha \in A$, $\, V_\alpha$ is open.
Indeed, as $F$ discrete, $U\times \{\alpha\}$ is open in $U \times F$ whence, by continuity of $h$, $V_\alpha$ is open in $p^{-1}(U)$.
It remains to prove that, for each $\alpha \in A$, $\; p|_{V_\alpha}$ is a homeomorphism.
Let $\alpha \in F$ be given. Then, by hypothesis, $p|_{V_\alpha}=\pi_1\circ h \circ \iota, \,$ where $\iota:V_\alpha \hookrightarrow p^{-1}(U)$.
Therefore, since $V_\alpha:=h^{-1}(U\times \{\alpha\})$, we have $p|_{V_\alpha}=\pi_1|_{U\times \{\alpha\}}, \,$ clearly a homeomorphism.
