Is the extension of a completely regular topology to a larger set completely regular?

Question

Let

1. $(X,\mathcal T)$ be a completely regular topological space (not necessarily Hausdorff)

2. $Y$ be any superset of $X$. (you can suppose $|Y\setminus X|=1$).

3. $\mathcal S$ be the topology on $Y$ generated by $\mathcal T$.

Is $(Y,\mathcal S)$ completely regular?

Is $X$ dense in $Y$?

Answer 1

It will never be regular, let alone completely regular. Note that the topology $\mathcal{S}$ will be $\mathcal{T} \cup \{ Y \}$. It follows that given any nonempty $A \subseteq Y$ we have $Y \setminus X \subseteq \overline{A}$. Therefore if $x \in X$ and $U$ is an open neighbourhood of $x$ in the $\mathcal{T}$-topology, then for any open neighbourhood $V$ of $x$ we will have $\overline{V} \nsubseteq U$.

$X$ will be dense in $Y$, for the reason noted above.