Is the closure of a Hausdorff space, Hausdorff?

For a generalisation, one can easily show that every Hausdorff space can be embedded as a dense subset of a non-Hausdorff space. Given a Hausdorff space $X$, define a new space $Y = X \cup \{ \alpha , \omega \}$ so that the open sets are just the open sets of $X$ together with the whole space $Y$. Then $X$ is clearly a dense subspace of $Y$ (the only neighbourhood of either $\alpha$ or $\omega$ is the entire space $Y$), and $Y$ is not Hausdorff (actually, it is not even T$_0$, since $\alpha$ and $\omega$ have exactly the same open neighbourhoods).

No.

Consider $\Bbb R$ with the trivial topology $\{0\}$ is a dense subset, it is certainly Hausdorff, but space itself is certainly not.

For a slightly less trivial example, consider $X = [0,1] \cup \{1'\}$, where $1'$ is an "extra copy of $1$" so that basic neighbourhoods of $1'$ are $\{1'\} \cup (1-\epsilon,1)$ for any $\epsilon > 0$. On $[0,1]$, the usual topology is taken. Then $[0,1]$ is a dense subspace which is Hausdorff, but $X$ is not Hausdorff because $1$ and $1'$ don't have disjoint neighbourhoods.