To prove that $(\mathbb{P(N)},d)$ is a metric space where $\mathbb{P(N)}$ is the power set of $\mathbb{N}$. And $d:\mathbb{P(N)}\times \mathbb{P(N)} \mapsto\mathbb{R}$ is defined by $d(A,B)=\begin{cases} 0, & \text{if A} \Delta \text{B=}\emptyset \\ m^{-1}, & \text{where m is the smallest member of} A\Delta B \end{cases} $.
Attempt: $x \in P\Delta Q=(P-Q)\cup(Q-P) \implies x\in (P-Q) \text{ or } x\in (Q-P) $. But $(P-Q)\cap(Q-P)= \emptyset \implies x \in P \text{ and} \notin Q \text{ or }x \in Q \text{ and} \notin P. \ \ \ $ (*)
To prove the triangle inequality:
Take any three sets $A,B,C \subset \mathbb{P(N)}$ such that none of them are equal to another. Now,
$\min A \Delta B =m_1, \min B \Delta C =m_2, \min A \Delta C =m_3 $. To prove $d(A,B) \leq d(A,C)+ d(B,C)$. By (*), $m_1 \in A$ or $m_1 \in B$ but not both.
Again, $m_2 \leq m_1$ or $m_3 \leq m_1$. Either way, we are done. When $A \neq B$ and either $B =C $ or $A=C$ (not both), we have respectively $A=C$ , $B=C$ . Which satisfies the triangle inequality. When $A=B$ it is evident.
Symmetry: We know $A \Delta B =B \Delta A$. $d(A,B)= d(B,A)$
Non-negativity and $0 \iff A=B$:
$d(A,B)=0 \iff A \Delta B= \emptyset \iff A \subset B \text{ and } B \subset A \iff A=B$. By definition $d(A,B) \geq 0$
Is this okay? Please verify.
Use that $A \Delta C\subseteq (A \Delta B) \cup (B \Delta C)$, so $\min(A\Delta C) \ge \min(\min(A \Delta B), \min(B \Delta C))$ and so (as $m \to \frac1m$ reverses order) $d(A,C) \le \max(d(A,B), d(B,C))$ which shows that $d$ is even an ultrametric.
I don't see why $m_2 \le m_1$ or $m_3 \le m_1$ is justified, you give no argument for that. I concur symmetry and the remaining axioms are all trivial.