Let $X$ be a perfect Polish space and let $H[X]$ be the set of all non-empty compact subsets of $X$. For $A,B \in H[X]$ define the so called Hausdorff-Distance $$ d_H(A,B) = \max \{ \sup_{x \in X} \inf_{y\in Y} d(x,y), \sup_{y\in Y} \inf_{x \in X} d(x,y) \} $$ which is a metric on $H[X]$, proof that $H[X]$ is perfect.
I have to show that each open ball $B_{r}(A) = \{ B \in H[X] : d_H(A,B) < r \}$ contains a point $B \ne A$. For this I firstly wanted to show that the set (the set of points which have a distance $<r$ from $Y$) $$ Y_r = \bigcup_{y \in Y} \{ x \in X : d(x,y) < r \} $$ contains a point not in $Y$. But I am unable to do. I have that $Y$ is compact, and because I am in a metric space so this is equivalent with $Y$ being closed and totally bounded. Guess it would be enough to show that the boundary of $Y$ is non-empty. So I tried, using totally bounded, to construct a sequence which approaches a point on the boundary, and then using closeness to deduce that it lies in $Y$, but I have now idea how to find such a sequence? Any hints? Btw my other post If a set could be represented as "arbitrary fine" finite union of open balls, then it is not closed is related to this question, but after the answers I don't have any ideas to proceed.
Sometimes it helps to consider some simple special cases. Let's for simplicity take $X = [0, 1]$, and consider two extreme instances of $A$.
You will find that every other set is sufficiently similar to at least one of these cases.