Open image of a normal space that is not $T_2$ I'm trying to find an open surjective map $f:X\to Y$ such that $X$ is normal and $Y$ is $T_1$ but not $T_2$.
The truth is that the only space $T_1$ that is not $T_2$ that I know is $(X,\tau_c)$ where $X$ is infinite and $\tau_c$ the cofinite topology. 
Which normal space $X$ and $f$ could we take? Can anyone give me a hint?
Thank you.
 A: Corrected: Let $X=[0,1]$ with the following topology: points of $(0,1)$ are isolated, sets of the form $[0,x)\setminus C$ for $x\in(0,1]$ and countable $C\subseteq(0,x)$ form a local base at $0$, and sets of the form $(x,1]\setminus C$ for $x\in[0,1)$ and countable $C\subseteq(x,1)$ form a local base at $1$. Define an equivalence relation $\sim$ on $X$ as follows: $x\sim y$ if and only if $x=y$, or $x,y\in(0,1)$ and $|x-y|\in\Bbb Q$. The $\sim$-equivalence class of $x\in[0,1]$ is $\{x\}$ if $x=0$ or $x=1$, and it’s $\{y\in(0,1):|y-x|\in\Bbb Q\}$ otherwise. Clearly $X$ is $T_3$.
Suppose that $H$ and $K$ are disjoint closed sets in $X$. If $H\cup K\subseteq(0,1)$, then $H$ and $K$ are clopen, so they are disjoint open sets separating themselves. If $0\in H$, there are an $x\in(0,1)$ and a countable $C\subseteq(0,x)$ such that $\big([0,x)\setminus C\big)\cap K=\varnothing$. If $1\notin H$, let $U=H\cup\big([0,x)\setminus C\big)$; then $U$ is a clopen nbhd of $H$ disjoint from $K$, so $U$ and $X\setminus U$ are disjoint open sets separating $H$ and $K$. The other possibilities are handled similarly, and we see that $X$ is normal.
Now let $Y=X/\!\!\sim$ be the quotient space, and let $f:X\to Y$ be the quotient map. Let $p=f(0)$, $q=f(1)$, and $Y_0=Y\setminus\{p,q\}=f\big[(0,1)\big]$. Each point of $Y_0$ is an isolated point of $Y$. For any $x,y\in[0,1]$ with $x<y$, the interval $(x,y)$ contains a point from each of the infinite $\sim$-classes, so in particular $f\big[[0,x)\big]=\{p\}\cup Y_0$ for each $x\in(0,1]$, and $f\big[(x,1]\big]=\{q\}\cup Y_0$ for each $x\in[0,1)$. Thus, if $C\subseteq(0,1)$ is countable, then $$f\big[[0,x)\setminus C\big]=\{p\}\cup(Y_0\setminus D)$$ and $$f\big[(x,1]\setminus C\big]=\{q\}\cup(Y_0\setminus D)$$ for some countable $D\subseteq Y_0$. It’s easily checked that the inverse images of these sets under $f$ are open in $X$, so $f$ is an open map.
We’ve just seen that basic open nbhds of $p$ in $Y$ have the form $\{p\}\cup(Y_0\setminus C)$ for some countable $C\subseteq Y_0$, and basic open nbhds of $q$ have the form $\{q\}\cup(Y_0\setminus C)$ for some countable $C\subseteq Y_0$. It follows immediately that $Y$ is $T_1$. However, $Y_0$ is uncountable, so $U\cap V\ne\varnothing$ whenever $U$ and $V$ are open nbhds of $p$ and $q$, respectively, so $Y$ is not Hausdorff.
