Every DFA is an NFA

Define an NFA $N=(Q,\mathrm{\Sigma },\eta ,q_0,F)$ by $\eta (q,a):=\{\delta (q,a)\}$ for $a\in \mathrm{\Sigma }$ and $\eta (q,\epsilon ):=\mathrm{\varnothing }$ for every state $q$. Step [L1] makes this well-defined, and $N$ has the same states, start state, and accepting states as $D$.

We claim that for every state $q$ and every word $w$, the NFA reachable set from $q$ on $w$ is the singleton containing the DFA state reached from $q$ on $w$. This is proved by induction on $\mathrm{\mid }w\mathrm{\mid }$: for $\epsilon$ there are no $\epsilon$-moves, so the reachable set is $\{q\}$; for $wa$, the NFA applies the same one-letter transition as the DFA and then again has no $\epsilon$-moves.

By step 2.1 and [L2], a word $w$ reaches an accepting DFA state exactly when the NFA reachable set on $w$ contains an accepting state. By [L3], this is exactly the NFA acceptance condition, so $L(N)=L(D)$.