Proving that quotient space is Hausdorff Let $(X, \tau_X)$ a compact space, $\simeq$ a equivalence relation on $X$ and $p:(X, \tau) \to (X/\simeq, \tau_\simeq)$ defined by $p(x)=[x]$. Prove if $p$ is closed then $ (X/\simeq, \tau_\simeq)$ is Hausdorff.
Since $X$ is compact and $p$ is continuous and surjective we have that $ (X/\simeq, \tau_\simeq)$ is compact. I need ideas to continue.
 A: Note I use the notation $[x]$ to denote the equivalence class of $x$ both as a subset of $X$ and as an element of $X/\simeq$.  I hope it is clear from context which I mean at any given time, but let me know if not.
Firstly, we must assume that $(X, \tau_X)$ is Hausdorff.  Otherwise simply take the trivial relation: $x\simeq y \iff x=y$, to get a contradiction.
Secondly we must unpack the meaning of $p$ being closed.  This says that for any open set $U\subseteq X$, the image under $p$ of the complement of $U$ is closed:
$$\{[y] | \exists x\notin U, x\simeq y\} \qquad \text{is closed.}$$
Equivalently, we may state this by saying the complement of this set is open:
$$\{[z] | w\simeq z\implies w\in U \} \qquad \text{is open.}$$
Finally, we note that a set in $X/\simeq$ is open if and only if its preimage in  $X$ is open.  Thus $p$ is closed means that for every open set $U\subseteq X$, we have:
$$\hat{U}=\{z\in X | w\simeq z\implies w\in U \} \qquad \text{is open.}$$
Here, for any open set $U$, we have defined $\hat{U}\subseteq U$ to be the set of elements whose entire equivalence class is contained in $U$.  Then we have shown that $p$ being closed is equivalent to the statement: $\hat{U}$ is open, for every open $U\subseteq X$.
Now lets use the fact that $X$ is compact to show that any open cover of an equivalence class has a finite subcover.  First note that as $X$ is Hausdorff, points are closed: $\{x\}^C$ is open for all $x\in X$. Thus $\hat{\{x\}^C}$ is open.  However $\hat{\{x\}^C}=[x]^C$, the complement of the equivalence class $[x]$.
Thus given any open cover of $[x]$, we may add the open set $[x]^C$, to get a cover of $X$, take a finite subcover and discard $[x]^C$, to get a finite subcover of $[x]$.
We are now ready to begin. Given $[x]\neq [y]$ we will find disjoint neighbourhoods of $[x]$ and $[y]$ in $X/\simeq$.
Let $z\simeq y$.  For every $w\simeq x$ we may take disjoint neighbourhoods $U_w, V_w$ of $w,z$ respectively.  A finite number of the $U_w$ cover $[x]$: $U_{w_1},\cdots, U_{w_n}$.  Let $U_z$ be the union of the $U_{w_i}$ and let $V_z$ be the intersection of the $V_{w_i}$.
Now $U_z,V_z$ are disjoint neighbourhoods of $[x], z$ respectively.  If we do this for each $z\simeq y$, then we have a finite subcover $V_{z_1},\cdots,V_{z_m}$ of $[y]$.  Now let $U$ be the intersection of the $U_{z_j}$ and let $V$ be the union of the $V_{z_j}$.
We now have $U, V$ disjoint neighbourhoods of the subsets $[x],[y]$ of $X$.  Then $p(\hat{U}), p(\hat{V})$ are disjoint neighbourhoods of $[x],[y]\in X/\simeq$.
