Let $f:X\ \to\ Y$ be a function. Recall that $f$ induces a function $f:P(X)\ \to\ P(Y)$. To prove that $f[\emptyset]\ =\ \emptyset$. We usually prove the two inclusions: $\ \ \emptyset\ \subseteq\ f[\emptyset]\ \hbox{and}\ f[\emptyset]\ \subseteq\ \emptyset$.
The first inclusion $\emptyset\ \subseteq\ f[\emptyset]$ is always true. Then it suffices only to prove the inclusion $f[\emptyset]\ \subseteq\ \emptyset$. To do this, we must show that the implication $\ \ z\ \in\ f[\emptyset]\ \longrightarrow\ z\ \in\ \emptyset$ is not false for an arbitrary $z$.
The above implication is false if and only if its antecedent $\ z\ \in\ f[\emptyset]\ $ is true and its consequent $z\ \in\ \emptyset$ is false.
But the consequent $\ z\ \in\ \emptyset\ $ is clearly false and so the implication $z\ \in\ f[\emptyset]\ \longrightarrow\ z\ \in\ \emptyset$ can only be true if its antecedent $z\ \in\ f[\emptyset] $ is false.
But $z\ \in\ f[\emptyset]$ is false only if there are no elements in $f[\emptyset]$; consequently, this means that the set $f[\emptyset]$ must be empty; and so $f[\emptyset]\ =\ \emptyset$.