A problem of metrization Let $X$ be a compact metric space and $f:X \to Y$ continuous surjective map. 
I need prove that $Y$ is metrizable.
Can I prove that $Y$ is Hausdorff?
Because if $Y$ is Hausdorff, i can prove $Y$ is metrizable by a Willard's theorem.
I tried to prove that $f$ is closed to use another result and have that $Y$ is Hausdorff, but I think that is not true.
What could I use it to see that Y is metrizable?

This is false. Let $X$ be any compact metric space with at least two elements (say $X=[0,1]$), let $Y$ be $X$ endowed with the trivial topology, and let $f$ be the identity map. Then $f$ is a continuous surjection, but $Y$ is not Hausdorff and thus not metrizable.
