Closed subsets of metrizable compact spaces can be viewed as intersection of closed with non-empty interior Let $X$ be a compact, metrizable space, and let $Y\subseteq X$ be a closed subset.
Is is always possible to find a decreasing sequence $(Y_m)_{m\geq 1}$ of closed subsets of $X$, s.t. $\mathrm{int}(Y_m)\neq\emptyset$ and $Y=\bigcap\limits_{m=1}^{\infty} Y_m$?
If $Y$ is a subset of finitely many points, I know that the answer is yes.
I also know that if $X$ is not compact, this fails.
 A: For every integer $n$, there exists a finite number of elements $y_{i_1},..,y_{i_n}$ such that $y_{i_j}\in Y$ and $Y_n=B(y_{i_1},{1\over n})\cup...\cup B(y_{i_n},{1\over n})$ contains $Y$ since $Y$ is compact.
$Z=\cap_{n>0}Y_n=Y$. Suppose this is not true. There exists $z\in Z$ not in $Y$. For every $n>0$, $z\in B(y_{i^z_n},{1\over n})$. Since $Y$ is compact we can extract a sequence from $(y_{i^z_n})$ which converges towards $y\in Y$. $d(y,z)\leq d(y,y_{i^z_n})+d(y_{i^z_n},z)$. This implies that $y=z$ contradiction.
To make the sequence $Y_n$ a decreasing sequence, replace it by $Z_n=\cap_{i=1}^{i=n}Y_i$.
A: Counterexample: 
Space $X$ has a finite underlying set and is equipped with discrete topology. 
Let $Y=\varnothing$. 
Then by decreasing $Y_m$ the equality $\bigcap_{m=1}^{\infty}Y_n=Y=\varnothing$ implies that $Y_m=\varnothing$ for $m$ large enough.
Then also $\mathsf{int}(Y_m)=\varnothing$.
A: $Y=\emptyset$ is a counterexample in any compact space, since the intersection of a decreasing sequence of nonempty closed sets is nonempty.
If we assume $Y\ne\emptyset$ then compactness is not needed: if $Y$ is a nonempty closed set in a metric space $X$, then $Y$ is the intersection of a decreasing sequence of closed sets with nonempty interiors.
Proof. Choose a point $p\in Y$. Let $Y_n=Y\cup\{x\in X:d(x,p)\le\frac1n\}$. Then $Y_n$ is a closed set with nonempty interior (since $p$ is an interior point of $Y_n$), and $\bigcap_{n=1}^\infty Y_n=Y$.
That is a rather trivial and uninteresting statement. Here's a better one:
Proposition. In a metric space $X$, any closed set $Y$ is the intersection of a decreasing sequence of closed sets $Y_n$ such that $Y\subseteq\operatorname{int}(Y_n)$.
Proof. Let $Y_n=\{x\in X:d(x,Y)\le\frac1n\}$. Then $\bigcap_{n=1}^\infty Y_n=Y$ because $Y$ is closed, and $Y\subseteq\{x\in X:d(x,Y)\lt\frac1n\}\subseteq\operatorname{int}(Y_n)$.
A: The conclusion holds in any metric space. Compactness is not necessary. (I will assume that $Y$ is non-empty). Let $W_n =\{x: d(x,Y) <\frac 1 n\}$. Let $Y_n=\overset {-} {W_n}$. Then $W_n$ is a nonempty open set for each $n$ , so $Y_n$ is  a closed set with non-empty interior. Since $Y_n \subset \{x: d(x,Y) \leq \frac 1 n\}$ it is clear that $\cap_n Y_n =Y$. 
Any point of $Y$ is an interior point of $W_n$ because it belongs to $W_n$ and $W_n$ is open (by continuity of the map $ x \to d(x,Y)$). 
