$A$ is closed $\implies X/A$ is Hausdorff whenever $X$ is Hausdorff. 
Let $X$ be a Hausdorff topological space. If $A$ is a closed subset of $X$ then $X/A$ is Hausdorff.

Let $\{x\},\{y\} \in X/A$ such that $x,y \in X \setminus A$ with $x \neq y.$ Then since $X \setminus A$ is open in $X$ and since any subspace of a Hausdorff space is Hausdorff there exist open neighbourhoods $U_x$ and $U_y$ of $x$ and $y$ respectively in $X$ such that $U_x, U_y \subseteq X \setminus A$ and $U_x \cap U_y = \varnothing.$ But then $U_x$ and $U_y$ are saturated open subsets of $X$ with respect to $p.$ So $p(U_x)$ and $p(U_y)$ together gives a separation of $\{x\}$ and $\{y\}$ in $X/A.$
Now if I take $x \in X \setminus A$ then how do I separate $\{x\}$ and $A$ in $X/A\ $? Inspite of my tireless effort I am unable to show that. Would anybody please give me some suggestion which enables me to prove the rest of the result?
Thanks for your time.
EDIT $:$ Here $p : X \longrightarrow X/A$ is the quotient map.
 A: This would work if $X$ is, additionally, regular. Seperating the point $A$ from $\{x\}$ in $X/A$ is essentially the same as seperating the closed subset $A$ from the point $x$ in $X$, but this is not implied by the Hausdorff property - it is called regularity.
To understand this, let's think about what an open neighborhood of $A$ looks like in $X/A$. By definition, a neighborhood $A \in U \subseteq X/A$ is open iff $p^{-1}[U]$ is open in $X$. The preimage of $\{A\}$ is $A$, and the preimage of any other point $\{x\}$ is just $\{x\}$. So taking the preimage of our neighborhood $U$ has the effect of replacing the point $A \in X/A$ with the subset $A \subseteq X$. Thus, open neighborhoods of $A$ in $X/A$ are precisely those subsets which pull back to open subsets of $X$ containing $A$.
Now, suppose indeed that we could find disjoint open neighborhoods $A \in U$ and $\{x\} \in V$. Then we'd have $A \subseteq p^{-1}[U]$ and $x \in p^{-1}[V]$. But then these preimages are disjoint open subsets of $X$ which separate the subset $A$ from the point $x$. Thus, if the proposition you wrote was true then $X$ would have to be regular. Unfortunately, not all Hausdorff spaces are regular. See here for an example.
