Proposition 2

Let $A$ be a dense subset of a topological space $X$; then for each $x \in A$ and each neighbourhood $V$ of $x$ relative to $A$, the closure $\overline{V}$ of $V$ in $X$ is a neighbourhood of $x$ in $X$.

For $V$ contains $U \cap A$, where $U$ is an open subset of $X$ which contains $x$, hence $\overline{V}$ contains $U \cap \overline{A} = U$ (§ 1, no. 6, Proposition 5).