Proving this quotient space is a Hausdorff space 
Define $S^1 = \{x \in \mathbb{R}^2 : x^2 + y^2 = 1 \}$. Define the equivalence relation $\sim$ as follows: $(x,y) \sim (x',y')$ if and only if $y = y'$. Now prove that the quotientspace $X/\sim$ with the quotient topology is a Hausdorff space.

My approach:
Define $q: S^1 \rightarrow X/\sim$ as the canonical map, then $q$ is a quotient map. We know that $S^1$ is a Hausdorff space, since $S^1 \subset R^2$. Now let $x,y \in X/\sim$, with $x\neq y$. We have $q^{-1}(x) \neq q^{-1}(x)$. Due to the Hausdorff property of $S^1$ there exists open subsets $U, V$ such that $q^{-1}(x) \in U$ and $q^{-1}(y) \in V$ with $U \cap V = \emptyset$. 
Now I would like to say that $q(U)$ and $q(V)$ are the open subsets such that $x \in q(U)$ and $y \in q(V)$ and $q(U) \cap q(V) = \emptyset$. However, $q(U)$ is open iff $q^{-1}(q(U))$ is open and I'm not sure if I can just say $q^{-1}(q(U)) = U$. 
The question: am I on the correct path and how can I finish this?
 A: Note that $(x,y)\sim(x',y')$ implies $$x^2+y^2 = (x')^2 + y^2 = 1, $$ from which $x^2 =(x')^2$. Let $\pi:S^1\to S^1/\sim$ be the canonical projection map, then $$\pi(x,y) = \{(x,y),(-x,y) \}. $$ Now if $\pi(x,y)\ne \pi(x',y')$, we have $y\ne y'$, so as $S^1$ is Hausdorff as a closed subspace of $\mathbb R^2$, there exist disjoint neighborhoods $U, U', V, V'$ of $(x,y), (x',y'), (-x,y), (-x,y')$ in $S^1$. By definition of quotient topology, $\pi(U\cup V)$ and $\pi(U'\cup V')$ are neighborhoods of $\pi(x,y), \pi(x,y')$, and these are disjoint by construction. It follows that $S^1/\sim$ is Hausdorff.
A: Take the diagonal $\;\Delta\;$ in the quotient space:
$$\Delta=\left\{\,\left((x,y),\,(x,y)\right)\in X/\sim\,\right\}$$
and suppose we have a sequence $\;\left\{\,((x_n,y_n),(x_n,y_n))\,\right\}\subset\Delta\;$ that converges in $\;X/\sim\;$. We want to prove the limit is also in $\;\Delta\;$...but this is trivial since any set of representatives of the equivalence clases in $\;\Delta\;$ can be seen as elements in the diagonal of $\;\Bbb R^2\times\Bbb R^2\;$ , which is closed as the plane is Hasudorff, and thus the limit is also in the diagonal.
