Prove that if for every vertex in $G[U]$ there is an edge in to every vertex in $G[V-U]$ then there is a path from every $u$ to every $v$
You can use a direct proof. There is no need to assume that the graph is connected.
Take any $u \in U$. We will construct a directed path $u_0\to\cdots\to u_\ell$ in which $u_\ell \in U$ has out-degree zero in $G[U]$, as follows. First, $u_0 = u$. Given $u_i$, if $u_i$'s out-degree in $G[U]$ is zero, then $\ell = i$. Otherwise, let $u_{i+1}$ be an arbitrary out-neighbor of $u_i$. Since $G$ is acyclic, the vertices $u_i$ do not repeat, and so the path must terminate eventually.
Similarly, for any $v \in V \setminus U$ there is a directed path $v = v_0 \gets \cdots \gets v_r$, where $v_r \in V \setminus U$ has in-degree zero in $G[V \setminus U]$. By assumption, there is an edge $v_\ell \to u_r$. Combining both paths with this edge, we obtain a path from $u$ to $v$.