Chan,
Lets's say we have a DFA $D$ and we want to construct a NFA $N$, such that $L(D) = L(N)$. Then the construction would be as follows:
Let's assume that the DFA $D$ has the state set $Q = \{q_0, q_1, ..., q_n\}$.
Now, we "build" the NFA $N$ as follows:
($1$.) Start with the DFA $D$.
($2$.) Add an additional accepting state for the NFA $N$, such that $N$ will have $n + 1$ total number of states.
Let's call the new accepting state $q_{n + 1}$.
($3$.) Now, add an epsilon $\epsilon$ transition from all accepting states to the new accepting state $q_{n + 1}$, and make all the original accepting states just normal states.
DONE.
Now, we have an NFA $N$ such that $L(N) = L(D)$.
Note: Any of the states in the set $Q$ could be an accepting state. By making them regular states and adding epsilon $\epsilon$ transitions from them to our created accepting state, we have successfully built an NFA. This construction can work for any DFA.
Additional note: One variation of the above construction would be to add an additional start state as well, and make an epsilon $\epsilon$ transition from the new start state to the original start state.