Let $(X,\tau)$ be a $T_1$ topological space and $Y\subset X$ a dense subspace which is completely metrizable. Are there any sufficient conditions to ensure that $(X,\tau)$ is Hausdorff using the known facts?

EDIT: Here is an example of such a topological space which isn't Hausdorf, as Valerio asked.

Take $(Y,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ banach spaces such that there exists a sequence $(x_n)_{n\in\mathbb{N}}$ that converges to $x_1$ with respect with $\|\cdot\|_1$ and $x_2$ with respect to $\|\cdot\|_2$ and $x_1\not=x_2$ (This norms exist). Then construct the normed vector space $(Y,\max\{\|\cdot\|_1,\|\cdot\|_2\})$; in this new normed vector space the sequence doesn't converge but is Cauchy, so we complete it to the banach space $(X,\|\cdot\|_3)$. Now construct a new topology $\tau=\{A\ |\ A$ is open in $(X,\|\cdot\|_3) \land A\cap Y$ is open in $(Y,\|\cdot\|_1)\}$. Now $(X,\tau)$ is an example of a topological space like the original problem. It is no Hausdorf because the sequence I define has two different limits, $x_1\in Y$ and $x_3\in X-Y$

I know it is a bit long and I didn't prove all the things I claimed to be true, but this is the exact space I was working on. I'm trying to check what pairs of norms in $Y$ produce that if a sequence converges with respect to both of them, then it converges to the same element in $Y$.