Example of a $T_2$ non metric topological space (if possible). I´m looking for an example of a metric topological space $X$
 and a continuous surjection $f:X\to Y$
where $Y$  is Hausdorff but not metric. 
I'm wondering because my professor and I were studying the quotient topology. We were on: $X$ metric and $X/-$ Hausdorff, then $X/-$ metric. So then we asked ourselves if there was an example like the one above, but we left it because we where gonna study other stuff. So I´m still wondering.
 A: Here is an example where $f$ is a quotient map.  Let $X=[0,1]\times\mathbb{N}$, and let $\sim$ be the equivalence relation on $X$ given by $(x,n)\sim(y,m)$ iff either $(x,n)=(y,m)$ or $x=y=0$.  Consider the quotient space $Y=X/{\sim}$ and the quotient map $f:X\to Y$.  Then $Y$ is Hausdorff but not metrizable, because it is not first-countable.  (To prove it is not first-countable, suppose you have countably many open neighborhoods of the equivalence class $[(0,0)]\in Y$, and use a diagonalization argument to construct a new open neighborhood that does not contain any of them.)
A: Let $Y$ be the (Tychonoff) product of an uncountable family of nontrivial Hausdorff spaces, e.g., two-point discrete spaces. $Y$ is Hausdorff because any product of Hausdorff spaces is Hausdorff. $Y$ is not metrizable because it is not first-countable. Let $X$ be a discrete space with the same cardinality as $Y$ and let $f:X\to Y$ be a surjection; then $X$ is metrizable and $f$ is continuous.
A: It's certainly not true that if $X$ is metric and $X / R$ is a quotient of $X$ (in the quotient topology) such that $X / R$ is Hausdorff, that $X / R$ is metrisable.
A classical example of this is taking $X$ as the reals (usual topology) and $R$ the equivalence relation such that $xRy$ iff $x = y$ or $x,y \in \mathbb{Z}$ (identifying the integers to a point). This is not first countable at the class of the integers, but is Hausdorff.  
A: If you don't want $f$ to be a quotient mapping, then you can easily construct a meta-example, in the spirit of @bof's example.
Let $Y$ be your favourite Hausdorff, non-metrisable space, and let $X$ be $Y$ with discrete topology. Then the "identity" mapping $X\to Y$ works. Of course, it will never be a quotient mapping.
As an aside, what you are contemplating is true if you assume in addition that $X$ is compact. This is because in this case, $X$ is second-countable and $f$ is a closed mapping. This implies that $f[X]$ is also second-countable (because the image of a basis of closed sets in $X$ forms a basis in $f[X]$) and compact, and hence metrisable.
