Is every open set in a subspace the intersection of the subspace with an open set "having the right closure"? Let $X$ be a Hausdorff paracompact topological space and $Y ⊂ X$ closed. By definition every open $V ⊂ Y$ is the intersection of $Y$ with an open set $U ⊂ X$.
However for such an $U$ we are not guaranteed that $\overline{V} = \overline{U} \cap Y$, where the closure of $V$ is supposed to be taken in $Y$. But is it maybe possible to always choose such an $U$ for a given $V$?
Is there any additional assumption on $X$ (e.g. manifold) or on $V$ that would ensure this?
 A: (In this answer I use "normal" to mean disjoint closed sets have disjoint neighborhoods, without requiring points to be closed.)
The condition you ask for holds for a topological space $X$ iff $X$ is completely normal (i.e., every open subspace of $X$ is normal, or equivalently every subspace of $X$ is normal).  In one direction, if $X$ is completely normal, let $A$ be the boundary of $V$ in $Y$ and consider the subspace $Z=X\setminus A$.  Note that $V$ and $Y\setminus\overline{V}$ are then disjoint closed subsets of $Z$, so by normality they have disjoint neighborhoods $U$ and $W$ in $Z$.  Since $Z$ is open in $X$, $U$ is also open in $X$, and satisfies $U\cap Y=V$ and $\overline{U}\subseteq X\setminus W$ is disjoint from $Y\setminus\overline{V}$, as desired.
Conversely, suppose $X$ is not completely normal, so there is some open subspace $Z\subseteq X$ that is not normal.  Let $A$ and $B$ be disjoint closed subsets of $Z$ that cannot be separated by open sets.  Let $Y=A\cup B\cup (X\setminus Z)$.  Then $Y$ is a closed subspace of $X$ and $A$ is open in $Y$, but any neighborhood of $A$ in $X$ must have closure that intersects $B$, since the same is true in $Z$.
In particular, any metrizable space (and thus any manifold) is completely normal.  Paracompactness is neither necessary nor sufficient for this (even restricting to Hausdorff spaces).  For instance, a compact Hausdorff space is always paracompact but may not be completely normal (consider a compactification of any non-normal completely regular space), and a completely normal Hausdorff space may not be paracompact (for instance, the ordinal $\omega_1$ is completely normal but not paracompact).
