 Solution

The Baire category theorem says that a countable intersection of open dense sets in a nonempty complete metric space is dense. Write $E=\{e_1,e_2,\dots \}$. Each singleton is closed in a metric space and has empty interior because there are no isolated points. Therefore every $X\setminus \{e_j\}$ is open and dense. Applying the theorem also to $G$ yieldswhich is dense in $X$. Thus every nonempty open subset meets the required difference, not merely the original dense set $G$.