If A is compact and $A/{\sim}$ is Hausdorff, then $X/{\sim}$ and $A/{\sim}$ are homeomorphic. 
Let $X$ be a topological space and let $\sim$ be an equivalence relation on $X$.
Call a subset $A \subset X$ $\mathit{full}$ if every equivalence class intersects $A$. 
If A is compact and $A/{\sim}$ is Hausdorff, then $X/{\sim}$ and $A/{\sim}$ are homeomorphic.

I came across this Lemma on the web, and I can't see why it is true.
The basic idea of the proof is to use the well-known fact:
A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.
So the proof goes as follows:
First, it shows that there exists a continuous bijection $f$ from $A/{\sim}$ to $X/{\sim}$.
And it is easy to show that $A/{\sim}$ is compact.
But I couldn't show that $X/{\sim}$ is Hausdorff.
Why it is  Hausdorff? and why the assumption that $A/{\sim}$ is Hausdorff is needed?
 A: The result is false without some restriction on the equivalence relation. Let $X=[0,1]$ with the usual topology, let $A=\{0,1\}$, and let $\sim$ be the equivalence relation on $X$ whose equivalence classes are $\{0\}$ and $(0,1]$. Then $A/\!\sim$ is homeomorphic to $A$, the discrete two-point space, but $X/\!\sim$ is homeomorphic to the Sierpiński space, which is not even $T_1$.
A: This is just my first impression only as I do not have time to work out
the details or check the validity of my claims. I assume that $A$ is full, otherwise the claim is obviously false (for, let $A$ be a singleton).
You may think about the commutative diagram (Sorry, I don't know how
to draw but I can describe...):
$i:A\rightarrow X$, the usual inclusion map $i(x)=x$
$q_{1}:X\rightarrow X/\sim$, the canonical quotient map, $q_{1}(x)=\{y\in X\mid x\sim y\}$
$q_{2}:$$A\rightarrow A/\sim$, the canonical quotient map, $q_{2}(x)=\{y\in A\mid x\sim y\}$
$j:A/\sim\rightarrow X/\sim$, the map defined by $j(\alpha)=\beta$
such that $\alpha\subseteq\beta$. (That is, given $\alpha\in A/\sim$,
choose $x\in\alpha$. Then define $\beta=q_{1}(x)$.
Note that $q_{1}\circ i=j\circ q_{2}$.

To show that $X/\sim$ is Hausdorff:
Let $\beta_{1},\beta_{2}\in X/\sim$ and $\beta_{1}\neq\beta_{2}$.
Since $A$ is full, there exist $x_{1},x_{2}\in A$ such that $x_{1}\in\beta_{1}$
and $x_{2}\in\beta_{2}$. Check that $q_{2}(x_{1})\neq q_{2}(x_{2})$.
Since $A/\sim$ is Hausdorff, there exist disjoint open sets $U,V$
such that $q_{2}(x_{2})\in U$ and $q_{2}(x_{2})\in V$. Think about $U, V$ and the commutative diagram...
