In the book says: Consider the cases $x=y$, $x \not =y$, separately. Use Axiom of extent.
Case 1: $(x=y)$
If $x=y$ by hypothesis $\{x\}=\{u,v\}$. Let $z\in \{u,v\}$, by hypothesis we have $z\in\{x\}$, then $u=x \land v=x$. Using idempotence rule: $$(u=x \land v=x) \lor (u=x \land v=x)$$ By hypothesis: $$(u=x \land v=y) \lor (u=y \land v=x)$$.
But, how i can prove the proposition with $x\not =y$ ?.
Thanks
Case 2: $x \not = y$
By hypothesis $\{x,y\}=\{u,v\}$, then: $$(x=u \lor x=v) \land (y=u \lor y=v)$$ Using distribution: $$[(x=u \lor x=v) \land y=u] \lor [(x=u \lor x=v) \land y=v]$$ $$(x=u \land y=u) \lor (x=v \land y=u) \lor (x=u \land y=v) \lor (x=v \land y=v)$$ $$(x=y) \lor (x=v \land y=u) \lor (x=u \land y=v) \lor (x=y)$$ $$(x=y) \lor (x=v \land y=u) \lor (x=u \land y=v)$$ But $x\not =y$, then by conjunction: $$[(x=y) \lor (x=v \land y=u) \lor (x=u \land y=v)] \land (x \not =y)$$ Using Modus Tollendo Ponens (MTP); $$(x=v \land y=u) \lor (x=u \land y=v)\blacksquare$$
Thereom: $$\{x,y\}=\{u,v\} \rightarrow \bigg[[x=u \wedge y=v] \vee [x=v \wedge y=u]\bigg]$$
Proof (by Contradiction): $$ \begin{align} (x \neq u \vee y\neq v)\wedge (x\neq v \vee y\neq u) \leftrightarrow &\\ \big ((x \neq u \vee y\neq v)\wedge x\neq v \big ) \vee \big((x \neq u \vee y\neq v)\wedge y\neq u\big)\leftrightarrow &\\ \big ((x \neq u \wedge x \neq v) \vee (y \neq v \wedge x \neq v)\big) \vee \big((x \neq u \wedge y \neq u) \vee (y \neq v \wedge y \neq u)\big)\leftrightarrow &\\ \big ( x \notin \{u,v\} \vee v \notin \{x,y\}\big) \vee \big ( u \notin \{x,y\} \vee y \notin \{u,v\}\big ) \end{align}$$ But by hypothesis we have that $\{x,y\}=\{u,v\}$ therefore:$$\bigg[\big ( x \notin \{x,y\} \vee v \notin \{u,v\} \big ) \vee \big (u \notin \{u,v\} \vee y \notin \{x,y\} \big)\bigg ] \text{(Absurd!!)}$$