Hausdorff space has Hausdorff quotient if a compact is identified to a point? Is it true that if $X$ is $T2$ and $a \sim b \Leftrightarrow a = b \vee a,b \in K$ where $K$ is a compact subset of $X$ then $X/\sim$ is Hausdorff? If so how to prove it.
The original question asks if it is true that if $L,K \subset X$ are compact then there exist $U,V\in \mathcal{T}$ such that $K\subset U , L\subset V, U\cap V = \emptyset $. This second question follows easily from the first assertion.
 A: Both imply each other, and so are equivalent.
The second one goes as follows: given $x\in L$ and $y\in K$ define $U_{x,y}$ and $V_{x,y}$ to be open disjoint neighbourhoods of $x$ and $y$ respectively. These exist since $X$ is Hausdorff.
Now the trick is as follows: for a fixed $x\in L$ consider $\{V_{x,y}\}_{y\in K}$ which is a covering of $K$. And thus it has a finite subcover $\{V_{x,1},\ldots,V_{x,n}\}$. Now consider the corresponding open neighbourhoods $\{U_{x,1},\ldots, U_{x,n}\}$ of $x\in L$ and let $U_x=U_{x,1}\cap\cdots\cap U_{x,n}$ while $V_x=V_{x_1}\cup\cdots\cup V_{x,n}$. Thus we found an open neighbourhood of $x$ that doesn't intersect an open neighbourhood of whole $K$. Note that $K\subseteq V_x$ for any $x\in L$.
Now all $\{U_x\}_{x\in L}$ cover $L$. And so it has a finite subcover $\{U_1,\ldots, U_m\}$. We now consider corresponding $V_1,\ldots, V_m$ and do the opposite: we take union $U=U_1\cup\cdots\cup U_m$ and the intersection $V=V_1\cap\cdots \cap V_m$.
This careful construction shows that both are open and disjoint neighbourhoods of $L$ and $K$ respectively.

Now the first one. Take $[x],[y]\in X/\sim$. Let $\pi:X\to X/\sim$ be the quotient map. You can manually verify the following property of $\pi$. For any subset $A\subseteq X$ we have
$$\pi^{-1}(\pi(A))=\begin{cases}
A\cup K&\text{if }A\cap K\neq\emptyset \\
A&\text{otherwise}
\end{cases}$$
In particular $\pi^{-1}(\pi(A))=A$ if $K\subseteq A$. This property will be important in determining openness. Recall that for quotient maps we have: $W$ is open in $X/\sim$ if and only if $\pi^{-1}(W)$ is open in $X$.
WLOG we have two situations:

*

*$x\not\in K$, $y\not\in K$. Since $K$ is compact, then $X\backslash K$ is open. And thus there are open disjoint neighbourhoods $U,V$ of $x,y$ respectively such that $U\cap K=\emptyset=V\cap K$. This implies that $U=\pi^{-1}(\pi(U))$ and thus $\pi(U)$ is open. And so is $\pi(V)$. These are clearly disjoint, because $\pi$ is injective over $X\backslash K$.

*$x\not\in K$, $y\in K$. Then $\pi^{-1}([y])=K$ while $\pi^{-1}([x])=\{x\}$. Both are compact and so we can apply what we proved previously. We have open disjoint neighbourhoods $U$ of $\{x\}$ and $V$ of $K$. Now $U$ is disjoint from $K$ while $V$ contains $K$. Therefore $\pi^{-1}(\pi(U))=U$ and $\pi^{-1}(\pi(V))=V$ as well. Meaning $\pi(U)$ and $\pi(V)$ are again open. These are disjoint as well, which I leave as an exercise.

