Sets with Minimal Dense Subsets I'm hoping this question can be answered over a broader class of topological spaces (like regular countable spaces, metrizable spaces, or normal spaces maybe) but I do not know how difficult such a general problem will be so the main focus will be on subsets of $\mathbb R^N$ equipped with the standard topology: 

For any closed set, $A$, let $\Omega_A=\{B\mid \overline B=A\}$. Call a set, $B$, a minimal dense subset of a closed set, $A$, if $B\in \Omega_A$ and $C\subsetneq B\implies C\not\in \Omega_A$. Call a set an M-set if it is closed and has a minimal dense subset. Which sets are M-sets? Which sets have unique minimal dense subsets?

In an essence, for a given closed set, $A$, a minimal dense subset of $A$ is a 'smallest' set that has $A$ as its closure.  
A few immediate facts are that a minimal dense subset of a set will always be a subset of it and that finite sets will always have minimal dense subsets equal to themselves (hence, unique minimal dense subsets). 
There are also many examples of countably infinite M-sets, like $\mathbb Z$ or $\{0\}\cup \left\{\frac{1}{n}\mid n\in \mathbb Z^+\right\}$ or, in any $\mathbb R^N$, $\{\underbrace{(n,n,\dots,n)}_N\mid n\in\mathbb Z\}$.  
I have had difficulty thinking of any countably infinite closed sets that would lack a minimal dense subset. A tempting argument to make is that, if $X$ is any countably infinite closed set, throw out all of its limit points and what's left will suffice as a minimal dense subset of $X$. 
But this argument is too barebones. It doesn't use the fact that $X$ is closed nor that $X$ is countably infinite, so if it really worked it would apply to spaces that lack one of those properties. But it doesn't - it fails, for instance, for $\mathbb Q$ (countably infinite, but not closed) and it fails for $[0,1]$ (closed, but not countably infinite). The issue in both of these cases is that all points of these sets are limit points. To make the 'remove the limit points' argument work, there would need to be some assurance that the set of isolated points of the closed set is 'big' enough to approach all limit points. But I have been unable to construct such an argument for countably infinite closed sets, though it feels as though it should be obvious. Or perhaps it really does not work.
Speaking of $[0,1]$ - it looks to me like there should not be very many uncountable M-sets, if there are any (again, just a hunch). Certainly, in any $\mathbb R^N$, the $N$-dimensional unit cube should not be an M-set, and I suspect more generally that closed connected sets can never be M-sets. It feels to me like this ties back to the idea that M-sets can be obtained by getting rid of limit points.   
More exotic examples, like the Cantor set, also do not seem to be M-sets. But, again, I have not been able to prove this.  
The fact that, in general, $\overline{\cap_{\alpha\in J} F_{\alpha}}\neq \cap_{\alpha\in J}\overline{F_{\alpha}}$ also renders fruitless some of the more instinctive lines of thinking. Like ordering the subsets of a closed set, $A$, that have closure equal to $A$ by inclusion and using Zorn's lemma to show that a minimal dense subset exists. Or saying that, if $A$ is an M-set, then we may merely take the intersection of all of $A$'s minimal dense subsets and thus get a unique minimal dense subset for $A$. At the same time, I have been unable to think of any closed sets that could have multiple minimal dense subsets.
One can also reframe the problem in terms of interiors instead of closures. Then, instead of looking for closed sets, $A$, with 'smallest' sets possessing closure equal to $A$, we'd be looking for open sets, $O$, with 'biggest' sets possessing interior equal to $A$. I think this way of looking at the problem is more helpful, but I have still not been able to make real progress using it. 
I have been having difficulty making much headway in the form of direct proofs thinking about this problem. I apologize if the content of the question is too bloated for what may be a relatively simple problem. 
(feel free to comment or edit for any corrections)
 A: If a subset $M$ has no isolated points and we're working in a somewhat decent space ($T_1$ at least), then if $D$ is dense in $M$ then $D\setminus \{d\}$ is dense too. So no minimal dense subset exists. Otherwise, I think the isolated points of $M$ must be in fact dense (to have a minimal dense subset) and this is clearly minimal as isolated points have to be in any dense subset. 
So my hypothesis is that for halfway decent spaces we have that $A$ is an $M$-set iff it has a non-empty dense set of isolated points. Ordinals in the order topology are examples as well as discrete subspaces.
A: Suppose $A$ is a topological space with no discrete dense subset. Let $B$ be any dense subset of $A$. $B$ cannot be discrete so there exists some $y\in B$ that is a limit point of $B$.
Let $a$ be any element of $A$ and $O$ be any neighborhood of $a$. $\overline{B}=A$ means that $O$ intersects $B$. If $O$ does not contain $y$, then it is immediate that $O$ also intersects $B-\{y\}$. If $O$ does contain $y$, then $O$ is a neighborhood of $y$ as well. Given then that $y$ is a limit point of $B$, $O\cap B$ necessarily contains some $x\neq y$, so $O$ still intersects $B-\{y\}$. From this, it follows that $\overline{B-\{y\}}=A$, which means $B$ cannot be minimal. It follows that $A$ has no minimal dense subset, i.e. $A$ is not an $M$-set.
Now suppose $A$ is a topological space with a discrete dense subset, $B$.
Let $C$ be any proper subset of $B$. There necessarily exists some $b\in B$ such that $b\not\in C$. As $B$ consists of only isolated points, there is some neighborhood, $O$ of $b$, such that $O\cap B=\{b\}$. It follows that $O\cap C\subset O\cap (B-\{b\})=\emptyset$, so $b\not\in \overline{C}$ meaning $\overline{C}\neq A$. So $B$ is a minimal dense subset of $A$, making $A$ an $M$-set.
Now, suppose $A$ is an $M$-set with more than one minimal dense subset. By the first couple of paragraphs $A$ must have a discrete dense subset, $B$. By the third and fourth paragraphs, $B$ is also a minimal dense subset of $A$.
Finally, in $T_1$ spaces minimal dense subsets of $M$-sets are unique. In fact, if $A$ is a $T_1$ $M$-set, then the only minimal dense subset of $A$ is the subset of $A$ consisting of all isolated points in $A$.
Suppose $A$ is a $T_1$ $M$-set. Let $I$ denote the set of all isolated points in $A$. We know that $A$ has a minimal dense discrete subset, $B$. If there was any $i\in I-B$, then it is easy to see that $i\not\in \overline{B}=A$, a contradiction. So $I-B$ is empty, i.e. $B\supset I$. We wish to show also that $B\subset I$. Suppose there is some $b\in B-I$. This means all neighborhoods of $b$ contain points aside from $b$. As $b$ belongs to the discrete subset, $B$, there exists a neighborhood of $b$, $O$, such that $O\cap B=\{b\}$. This neighborhood too must contain some point of $A$, $y$, aside from $b$. $y$ must necessarily lie outside $B$ (by the way $O$ has been defined). As $A=\overline{B}$, $y$ is necessarily a limit point of $B$. But $O$ (containing $y$) is a neighborhood of $y$. As $A$ is $T_1$ and $y$ is a limit point of $B$, this means $O$ contains at least one point, $c\in B$, aside from $b$. This contradicts our assumption that $O\cap B=\{b\}$. So $B-I$ is empty, i.e. $B\subset I$, giving us $B=I$ as desired.
