Since the map $\pi:X\to X/\sim$ is open, it's clear that the map $g:X^2\to (X/\sim)^2$ given by $g(x,y)=(\pi(x),\pi(y))$ is open. What we claim is that $g(X^2-\sim)=(X/\sim)^2-\Delta_{X/\sim}$. Indeed, if $x\nsim y$ then $\pi(x)\ne\pi(y)$ which tells us that $g\left(X^2-\sim\right)\subseteq (X/\sim)^2-\Delta_{X/\sim}$. That said, if $(\pi(x),\pi(y))\notin\Delta_{X/\sim}$ then $\pi(x)\ne \pi(y)$ so that $x\nsim y$ so that $(x,y)\in X^2-\sim$ and clearly $g(x,y)=(\pi(x),\pi(y))$. Thus, $g(X^2-\sim)=(X/\sim)^2-\Delta_{X/\sim}$ as claimed. But, since $X^2-\sim$ is open by assumption, and $g$ is an open map we have that $(X/\sim)^2-\Delta_{X/\sim}$ is open, and so $\Delta_{X/\sim}$ is closed. This gives us $T_2$ness.
Consider a closed interval with infinitely many copies of the origin.
Take the interval $([-3,3], \tau)$, with a base $B$ for the usual topology, and add to the set $[-3,3]$ infinitely many copies of $0$: $\{(0, n) : n\in \mathbb{N}\}$. To the base $B$, add in copies of a neighbourhood base of $0$ for each copy of $0$. For example, one new base for this topology is $B$ union the set of all sets of the form $(-r, r) - \{0\}\cup \{(0,n)\}$ (where $r$ is a positive real, and $n$ natural).
It's clear from the way the neighborhoods of the copies of the origin are defined that any continuous map to a Hausdorff space must send all the origins to the same point.
You can get an open cover of the line with infinite origins that has no finite subcover as follows:
Let your open cover contain $[-3, -1)$, $(1, 3]$, $(-2,2)$, and every open set of the form $(-2, 2) - \{0\}\cup \{(0,n)\}$.
EDIT:
To be very explicit, we create a new topological space $(X, \tau)$.
Let $X = [-3,3] \cup \{(0, n) : n\in \mathbb{N}\}$ (you can imagine these points as being `borrowed' from $\mathbb{R}^2$, the point is just that they are new points not usually on the interval $[-3,3]$).
We let $\tau$ contain every open set that belongs to the normal Euclidean topology on $[-3,3]$. We also let $\tau$ contain every set of the form $(-r, r) - \{0\}\cup \{(0,n)\}$, where $r$ is real, and $n$ is natural.
AND we let $\tau$ contain any arbitrary union or countable intersection of any kind of set mentioned so far.
Best Answer
For $x\neq y\in X_{H}\implies$ that $x,y$ are not related in $X$.
So there exist a pair $(Y,f)$ such that $Y$ is Hausdorff and $f$ is continuous from $X\to Y$ such that $f(x)\neq f(y)$.
So we take disjoint open nbds $U$, $V$ of $f(x),f(y)$ in $Y$.
Then $\bar{U}=\bar{f}^{-1}(U)$ and $\bar{V}=\bar{f}^{-1}(V)$ are open nbds of $x$ and $y$ in $X_{H}$ which are disjoint ,where $\bar{f}$ is such that $\bar{f}\circ \pi=f$