Because of Theorem [dt-001O], it suffices to show the second condition of continuity, as f is monotonic.
Because A is finite, any directed subset X \subseteq A is also finite, and therefore \bigsqcup X \in X. Because f is monotone, f(x) \sqsubseteq f (\bigsqcup X) for all x \in X.
The solution to Exercise [dt-001N] shows that \{f(x) \mid x \in X\} is directed and therefore contains its upper bound c. Because f is monotone, f(\bigsqcup X) must be c (else it would contradict the above).
Therefore \bigsqcup \{f(x) \mid x \in X\} = f (\bigsqcup X), as required.