Munkres Topology: Corollary 60.4 says that $p^{-1}(y)$ is a 2 point set. Why is this?
I understand $P^2$ to consist of equivalence classes of $S^2$ which are 2 point sets of antipodal points $\{x,-x\}, x \in S^2$ andd $p(x)=\{x,-x\}$. Therefore, "$y$" actually stands for a 2 point set $y=\{z,-z\}, z \in S^2$. So, $$p^{-1}(y)=p^{-1}(\{z,-z\})=\{z,-z\}$$
meaning
$$p^{-1}(y)=y?$$
Maybe my confusion above is like confusing the number $7$ with set $\{7\}$.
As you said, the equivalence class of $x \in S^2$ is the two-point set $[x] = \{ x,-x\}$. You have $P^2 = \{ [x] \mid x \in S^2 \}$ and $p : S^2 \to P^2, p(x) = [x] = \{ x,-x\}$. $P^2$ is given the quotient topology induced by $p$. This shows that for $y = [z]$ you get in fact $$p^{-1}(y) = \{ z,-z\} = y .$$ However, you must note that on the left hand side $y$ is a single point of $P^2$ and on the right hand side $y$ is a set of two points of $S^2$.
To see the difference, consider any function $f : A \to B$ between two sets $A,B$. For $D \subset B$ we have $$f^{-1}(D) = \{ a \in A \mid f(a) \in D \},$$ i.e. $f^{-1}$ is a function from the power set $\text{Pow}(B)$ of $B$ to the power set $\text{Pow}(A)$ of $A$.
For $b \in B$ we have $$f^{-1}(b) = f^{-1}(\{ b \}) = \{ a \in A \mid f(a) = b \} .$$ You may regard this as a function $f^{-1} : B \to \text{Pow}(A)$. This is precisely our situation. In fact,we consider $$p^{-1} : P^2 \to \text{Pow}(S^2) .$$ But now $P^2 \subset \text{Pow}(S^2)$ and $p^{-1}$ is nothing else than the inclusion $P^2 \hookrightarrow \text{Pow}(S^2)$.
This is a typical situatiuon if we have an equivalence relation $\sim$ on a set $A$. Let $B = A/\sim$ be the set of all equivalence classes $[a]$ of points $a \in A$ wit respect to $\sim$. We have $B \subset \text{Pow}(A)$. If $p : A \to B, p(a) = [a]$, then $p^{-1} : B\to \text{Pow}(A)$ is nothing else than the inclusion $B \hookrightarrow \text{Pow}(A)$.