The Problem: Let $p: S^2\to P^2$ be the quotient map from the $2$-sphere $S^2$ to its projective plane $P^2$. Then $S^2$ has a countable basis $\{U_n\}$ implies that $P^2$ has a countable basis $\{p(U_n)\}$.
Source: Topology, $\mathit{2^d}$ edition by James Munkres.
My Attempt:
i) Suppose $\{y, -y\}\in P^2$, then $p(y)=\{y, -y\}$. Let $U_N$ be the basis element in $\{U_n\}$ that contains $y$, then $\{y, -y\}\in p(U_N)$.
ii) Suppose $\{y, -y\}\in p(U_1)\cap p(U_2)$; then there exists some $y_1\in U_1$ and some $y_2\in U_2$ such that $p(y_1)=p(y_2)=\{y, -y\}$. We need to show that there exists some $p(U_3)$ such that $\{y, -y\}\ni p(U_3)\subseteq p(U_1)\cap p(U_2)$.
Case I: $y_1=y_2=y$; then let $U_3$ be the basis element of $S^2$ such that $y\ni U_3\subseteq U_1\cap U_2$. $p(U_3)$ would then be the desire basis element in $P^2$.
Case II: $y_1\neq y_2$. That is, $y\in U_1$ and $-y\in U_2$ (or $-y\in U_1$ and $y\in U_2$). Screeching Halt.
My Question: the issue is that in Case II if neither $y$ nor $-y$ is in $U_1\cap U_2$, then we can't find a $U_3\subseteq U_1\cap U_2$ that contains either $y$ or $-y$, thus $\{y, -y\}\ni p(U_3)\subseteq p(U_1)\cap p(U_2)$ can never hold for any $U_3$. What am I missing? Any help would be greatly appreciated.
To finish Case II in the question, note $-y\in U_2\implies y\in -U_2$, thus $y\in U_1\cap-U_2$. Let $U_3\subseteq U_1\cap-U_2$ such that $y\in U_3$ ($U_3$ exists since $U_1\cap-U_2$ is open), then $\{y, -y\}\in P(U_3)\subseteq P(U_1)\cap P(-U_2)=P(U_1)\cap P(U_2)$, as desired.