Showing that a set is residual in the metric space $(A,d_A)$ 
Let $D = [c,d]\times [c,d]\subseteq \mathbb{R}^2$ and let $A$ be the set of all closed subsets of $D$. For $a \in D$ and $B\in A,$ define $d(a,B) := \min\{d(a,b) | b\in B\},$ where the $d$ inside the min is the Euclidean metric on $\mathbb{R}^2$. This is defined as closed and bounded subsets of $\mathbb{R}^2$ are compact and the function $b\mapsto d(a,b)$ is continuous so the image of $D$ under this map attains its minimum value. For $B,C \in A,$ let $d_A(B,C) = \max\{\max_{b\in B} d(b,C), \max_{b\in B} d(c,B)\}.$


Show that the set $G := \{B\in A: B\text{ has no isolated points}\}$ is residual in $(A,d_A).$

To show that the set $G$ is residual, it suffices to show that $G^C$ is a countable union of nowhere dense sets (i.e. sets whose closures have empty interior). If $B\in G,$ then for every $b \in B$ and $r > 0, B^*(b,r) \cap B \neq \emptyset,$ where $B^*(b,r)$ denotes the punctured open ball centered at $b$ or radius $r$. I know that every complete metric space with no isolated points is uncountable, but I'm not sure if this is useful. I don't know how to define the nowhere dense sets for $G^C$ (I'd guess something like $A_k := \{B\in A : \exists b \in B, B^*(b, \frac{1}k)\cap B = \emptyset\}$).

Edit: Could someone justify the following claims in the answer below:


*

*(First (why?) question) If $B$ is disjoint from the closure of $U$ or $B$ contains more than one point in $U$, then $B$ is not in the closure of $A_U$?

I tried proving that if $B$ is disjoint from the closure of $U$, then $B$ is not in the closure of $A_U$ by taking a sequence $(C_n)$ of $A_U$ that converges to $B$ and showing that this must contradict the fact that $B$ is disjoint from the closure of $U$. I know that for each n, $|C_n \cap U| = 1,$ but I'm not sure how to use this to formally prove the required result.
As for the second part (where $|B\cap U| > 1$), I'm also unsure how to complete a formal proof. Suppose for a contradiction that $B\in \overline{A_U}.$ Choose a sequence $(C_n)\subseteq A_U, C_n\to B.$ Let $a$ be a point in $B\cap U.$ Since $C_n \to B, d_A(C_n, B)\to 0.$ For each $n \in \mathbb{N},$ let $c_n$ be the unique point in $ C_n\cap U$. I'm not sure if $c_n\to a$.


*If $B\cap U = \emptyset$ or $B\cap \overline{U} \neq \emptyset$ or $|B\cap U| = 1,$ then we can find $C$ with two points in $U$ that's arbitrarily close to $B.$
Again, I'm not even sure how to find the two points in $C\cap U.$
?
 A: You are on the right track for how to find the nowhere dense sets by considering balls that witness that points are isolated.  However, proving your proposed sets $A_k$ are nowhere dense is rather tricky because of the existential quantifier on $b$ in the definition.  This makes it difficult to prove that any particular open ball is disjoint from $A_k$ and thus difficult to say things about the closure of $A_k$.
Here is a slightly different characterization of having isolated points that I think is easier to use because it avoids referring to any specific point of the set.  A set $B\subseteq D$ has an isolated point iff there is an open set $U\subseteq D$ such that $B\cap U$ has exactly one point.  Can you use this characterization to find a countable collection of nowhere dense sets that cover all the sets $B\in A$ that have an isolated point?
More details are hidden below.

 We'd like to have a nowhere dense set $A_U$ consisting of those $B$ such that $B\cap U$ has exactly one point, for each open set $U\subseteq D$.  Of course, there are uncountably many different open sets $U$, so this doesn't quite work.  But there is a countable basis for the topology of $D$, and it suffices to check the condition on a basis of open sets.  So, we can use the sets $A_U$ where $U$ ranges over a countable basis for $D$.

 It remains to check that $A_U$ is actually nowhere dense.  This means, for any $B\in A$ and any $\epsilon>0$, you want to find $C\in A$ such that $d_A(B,C)<\epsilon$ and $C$ is not in the closure of $A_U$.  To do this, there are a few different cases to consider, based on how $B$ interacts with $U$.  For instance, if $B$ is disjoint from the closure of $U$, or if $B$ contains more than one point of $U$, then $B$ itself is not in the closure of $A_U$ (why?).  If $B$ contains exactly one point of $U$ or is disjoint from $U$ but intersects the closure of $U$, then you can find $C$ near $B$ which contains two points of $U$ (why?) and so is not contained in the closure of $A_U$.

