If $A\subset \Bbb R^n$ is a set such that for every finite, nonempty $B\subseteq A$, $A\setminus B$ is closed, show that $A$ is closed. 
Let $A\subseteq \Bbb R^n$ be a set such that for every finite, nonempty $B\subseteq A$, we have that $A\setminus B$ is closed. Is $A$ is closed?

I already proved that every point of $A$ is an isolated point. Thus, no point of $A$ is a limit point of $A$.
Then I separated the following $2$ cases:


*

*$A'= \emptyset$.

*$A'\neq \emptyset$.


If it's the first case, we're done.
If it's the second, let $y\in A'$. For each $r_n=\frac 1 n$, we have points $q_n\in A$ such that $q_n\neq y$ and $|q_n-y|<\frac 1 n$. This is a sequence $\{q_n\}\subseteq A$, with $q_n\to y$.
Now, let $B=\{b_1,\dots,b_K\}$ call the elements of $B$ which appear in the sequence $(q_n)$ $q_{B_1},\cdots,q_{B_C}$.
Now, as $\{q_n:n\in \Bbb N\}$ is infinite$^{(*)}$, we have that $\{q_n\}\setminus \{q_{B_1},\cdots, q_{B_C}\}$ is infinite.
Now, let $Q_n=q_{n+\max\{B_i:1\leq i \leq C\}}$. This is a subsequence of $q_n$, thus $Q_n\to y$, but $Q_n$ is contained in $A\setminus B$, thus $y$ is a limit point of $A\setminus B$ (I have to use the converse of what I did in the beggining to say this), and then $A\setminus B$ is not closed.

I believe this is correct (modulo proving two statements I used). Is there a simpler proof of this fact? As my proof seems unnecessarily complicated.
 A: If $A=\emptyset$ we are done.
Otherwise pick  $a\in A$ and let $B=\{a\}$. Then $A$ is the union of the two closed sets $A\setminus B$ and $B$, hence closed.
A: What is the error in this proof? 
(Just for fun!)
If $A$ is finite then it is closed and we are done. 
So let us consider $A$ infinite.
Recall: $A$ is closed iff it contains all of its limit points.
Consider any limit point $L$ of $A$. Our goal is to show that $L \in A$.
Pick a sequence of elements $a_1, a_2, a_3, \ldots$ in $A$ that converge to $L$.
Now take $B = \{a_1\}$. 
Observe $A - B = A - \{a_1\}$ contains the sequence $a_2, a_3, a_4,  \ldots$, which also converges to $L$. 
Since $A - B$ is closed, $L \in A - B \subset A$ as desired. QED.
A: If $A$ has no limit points we are done.  So let's assume $A$ has limit points.
Let $a$ be a limit point of $A$.  $B$ is finite so $D = \{d(a,b)|b \in B; b\ne a\}$ is a finite set.  All elements of $D$ are strictly greater than 0 so let $\delta$ be such $0<\delta < \min D$.  
Then for any $\epsilon > 0$ than $N_{\min(\delta, \epsilon)}(a) \subset N_{\epsilon}$ contains points of $A$ not equal to $a$ and $\min(\delta, \epsilon) \le d(a,b)$ for any $b \in B$ where $b\ne a$, the neighborhood contains points of $A/B$ not equal to $a$.  But as $A/B$ is closed $a \in A/B \subset B$.
So all limit points of $A$ are in A and A is closed.
