See http://www.jstor.org/stable/1971241?seq=6#page_scan_tab_contents, pages 598 and 599.
Suppose we have $\pi: \widetilde Y \rightarrow Y \rightarrow X\times \overline D \rightarrow \overline D$, where $\widetilde Y \rightarrow Y$ is the normalization (which is finite -- we are working with varieties). Suppose furthermore that every fiber is a (single) rational curve $\pi ^{-1}(d), d \in \overline D$.
We want to reach a contradiction. To do this, we require an isomorphism from a dense open subset of $\widetilde Y_d$ to its image in $Y_d$ (for any $d \in \overline D$). According to Mori, this is the case because the singular locus of $Y$ does not contain any fiber of $Y$ over $\overline D$.
Why does it not contain any fiber? How does this guarantee the isomorphism?