Let $\{V_\alpha\}$, $\{W_\beta\}$ be partitions of $p^{-1}(U)$.
Pick $\alpha$ and consider $V_\alpha$, choose $e \in V_\alpha$. Since $e \in p^{-1}(U) = \bigsqcup_\beta W_\beta$, there exists a unique $\beta$ such that $e \in W_\beta$.
We want to show that $V_\alpha = W_\beta$. Suppose by way of contradiction that $V_\alpha \neq W_\beta$. Then $V_\alpha \setminus W_\beta \neq \emptyset$. We know $V_\alpha \cap W_\beta \neq \emptyset$ since each contain $e$. Notice that $V_\alpha \cap W_\beta$ is open since it is the intersection of two open sets. Moreover $V_\alpha = \big( V_\alpha \cap W_\beta \big) \cup \big( V_\alpha \setminus W_\beta \big)$. And of course $\big( V_\alpha \cap W_\beta \big) \cap \big( V_\alpha \setminus W_\beta \big) = \emptyset$. If we can show that $V_\alpha \setminus W_\beta$ is open, then we have a separation of $V_\alpha$. This is a contradiction to our hypothesis that $U$ is connected since $V_\alpha$ is homeomorphic to $U$ and is this connected as well. Hence, we can conclude that $V_\alpha = W_\beta$ and thus the partition is unique.
Now why is $V_\alpha \setminus W_\beta$ open? If $e' \in V_\alpha \setminus W_\beta$, there exists $\beta_{e'} \neq \beta$ such that $e' \in W_{\beta_{e'}}$. Notice that $V_\alpha \setminus W_\beta = V_\alpha \cap \bigcup_{e' \in V_\alpha \setminus W_\beta} W_{\beta_{e'}}$ which is open because $\bigcup_{e' \in V_\alpha \setminus W_\beta} W_{\beta_{e'}}$ is open since it is the union of open sets, and the intersection of two open sets is open.