Let $\pi:\;S^2\to S^2/{\sim}$ be the projection map where the relation on $S^2$ is $a\sim b\iff a =\pm b$. I am trying to show $S^2/{\sim}$ is Hausdorff.
So take $\alpha,\beta\in S^2/{\sim}$ then $\pi^{-1}(\alpha)=\{\pm a\},\pi^{-1}(\beta)=\{\pm b\}$. Take $\varepsilon<\frac{1}{2}\min\left\{\|a-b\|, \|a+b\|\right\}$ then $A=B_\varepsilon(a)\cap S^2, B=B_\varepsilon(b)\cap S^2$ are disjoint open neighbourhoods of $a,b$ (and $-A,-B$ for $-a,-b$). Now $\pi(A)\cap\pi(B)=\emptyset$ since $\pi(u)=\pi(v)\iff u=\pm v$ and $A,B$ disjoint.
But I can't see how to properly show that $\pi(A),\pi(B)$ are open? I know that not all projection maps are open maps so it's not immediate...
By definition of the quotient topology, $\pi(A)$ is open iff $\pi ^{-1}(\pi(A))$ is open. But this is just $A \cup -A$, which is a union of two open sets. Similarly for $B$.