= Solution
The <Baire category theorem> says that a countable intersection of open dense sets in a nonempty <complete metric space> is dense. Write $E=\{e_1,e_2,\ldots\}$. 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 $G$ yields
$$
G\setminus E=G\cap\bigcap_{j\geq1}(X\setminus\{e_j\}),
$$
which is \b[dense in $X$]. Thus every nonempty open subset meets the required difference, not merely the original dense set $G$.
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