Let $(X,\tau)$ be a topological space and $Y$ be a non-empty subset. Suppose that $Y$ is dense in $(X,\tau)$ and that there exists a topology $\tau^{\star}$ on $Y$ which is strictly finer than the subspace topology induced by restriction of $\tau$. 


Does there exist a topology $\tau'$ on $X$ whose restriction to $Y$ is $\tau^{\star}$ but such that $Y$ is dense in $(X,\tau')$?



Extension of refined subspace topology
Let $\mathcal{U} = \tau \cup \tau^\star$, and let $\tau'$ be the unique minimal topology on $X$ containing $\mathcal{U}$. Since $\tau$ and $\tau^\star$ are topologies, they are closed under finite intersection; and since $\tau^\star$ is finer than the subspace topology on $Y$, the intersection of a set in $\tau$ with a set in $\tau^\star$ is again in $\tau^\star$. Thus $\mathcal{U}$ is closed under finite intersection. It follows that 
$$
\tau' = \{ \mbox{all unions of sets in $\mathcal{U}$}\}.
$$
Accordingly, every set $W$ in $\tau'$ can be written (nonuniquely) in the form $W = U \cup V$, 
where $U\in \tau$ and $V\in \tau^\star$.


Now if $x\in X$ and $x\in W\in \tau'$, write $W = U \cup V$ as above. If $x\in U$, then since $U\in \tau$ and $Y$ is $\tau$-dense in $X$, $U\cap Y \neq \varnothing$; if $x\in V$, then $V$ is a nonempty subset of $Y$. Taken together we see that $W \cap Y \neq \varnothing$, so $Y$ is $\tau'$-dense in $X$. 


