Suppose, $X$ is a Polish space, $Y$ is a Polish subspace of $X$. $\{U_n\}_{n \in \Bbb N}$ is a basis of open sets of $X$.
Let $A = \{ x \in \overline {Y} : \forall \epsilon \exists {n}(x \in U_n \land \operatorname{diam}{(Y \cap U_n)} < \epsilon) \}$
$\overline {Y}$ is the closure of $Y$. $\operatorname{diam}{(Y \cap U_n)}$ is the diameter of $Y \cap U_n$.
Why $A$ is different from $\overline {Y}$?
Added: The proof is from an online note (page 11)about descriptive set theory. Since it's posted free online by the author, I take the liberty to paste a screenshot of it for convenience.
Let's look at an example. Let $X = \{1, 1/2, 1/3, \dots, 0\}$ with the Euclidean topology, and take $Y = X \setminus \{0\}$. Then the discrete metric $d(x,y) = \delta_{xy}$ on $Y$ is complete and compatible with the subspace topology. Now $0 \in \bar{Y}$. However, if $U_n$ is any open neighborhood of 0 (in the topology of $X$), then it contains infinitely many points of $Y$; in particular, at least two points, so $\operatorname{diam}(U_n \cap Y) = 1$ (where the diameter is computed with respect to $d$). Therefore $0 \notin A$, and we see explicitly that $\bar{Y} \ne A$.