Why $e(C_i) = D_i$ is correct assumption? (FLP Impossibility 1985 - Lemma 3)
The paper says


  By an easy induction, there exist neighbors $C_0, C_1 \in \mathscr{C}$ such that $D_i = e(C_i)$ is $i$-valent, $i = 0, 1$


Here is a proof:

The set of configurations forms the nodes of a multidigraph in which the edges are labelled by events. $\mathscr{C}$ is the set of nodes reachable in any number of steps from $C$ while not following any edges labelled $e$. $\mathscr{D}$ is the set of nodes reachable from $\mathscr{C}$ by paths labelled $e$.

Let $i$ be the valence of $e(C)$: $e$ applies to $C$ by assumption; it is in $\mathscr{D}$ by definition; it is univalent by the assumption from paragraph 2. Let $D$ be a $(1-i)$-valent node in $\mathscr{D}$, which exists because paragraph 3 shows that $\mathscr{D}$ has both 0-valent and 1-valent configurations. By the definition of $\mathscr{D}$, there must be some node $P \in \mathscr{C}$ such that $D = e(P)$. By the definition of $\mathscr{C}$, there is a path from $C$ to $P$ consisting of nodes in $\mathscr{C}$.

The valences of $e(C)$ and $e(P)$ are unequal. For every node $V$ along the path from $C$ to $P$, $e(V) \in \mathscr{D}$, so every one is univalent. Thus, there must be two adjacent nodes along the path with different valences under $e$. These constitute $\{C_0, C_1\}$.