What is fishy about this proof of closure set being closed We use $Fr(A)$ and $Cl(A)$ to denote frontier of a set $A \subset R^n$ and its closure, closure being defined as the union $A \cup Fr(A)$. I am doing the exercise that says prove that $Cl(A)$ is a closed set.
Here is what I did.

By definition;
$Cl(A)$ closed $\iff $ $\forall \vec x\not \in Cl(A).\exists B(\vec x,r)$ such that $B(\vec x, r) \cap Cl(A) = \emptyset$
So, all points not in the set $Cl(A)$ are exterior points, if $Cl(A)$ is closed. Now, we are going to follow proof by contradiction. Assume closure of the set $A$, $Cl(A)$, is not closed.
$Cl(A)$ is not closed $\iff $ $\exists \vec x\not \in Cl(A)$ such that $\forall B(\vec x,r)$.$B(\vec x, r) \cap Cl(A) \neq \emptyset$
There is a point $\vec x_1 \not \in Cl(A)$ such that all balls around it intersects with $Cl(A)$. 
This $\vec x_1 \not \in Cl(A)$ implies $\vec x_1 \not \in A \cup Fr(A)$. Thus, $\vec x_1$ is also exterior to $A$. $\vec x_1$ being exterior to A, by definition, there should be a ball, say $B(\vec x_1, r_1)$, such that $B(\vec x_1, r_1) \cap A = \emptyset$.
But we said for this $\vec x_1$ that all balls around it intersects with the $Cl(A)$
$B(\vec x_1, r_1) \cap Cl(A) \neq \emptyset$ and $B(\vec x_1, r_1) \cap A = \emptyset$. $Cl(A) = A \cup Fr(A)$
So, $B(\vec x_1, r_1) \cap Fr(A) \neq \emptyset$ and  $B(\vec x_1, r_1) \cap A = \emptyset$
If, $B(\vec x_1, r_1)$ only intersects with the $Fr(A)$, than any smaller ball, like $B(\vec x_1, r_1 / 2)$, does not intersect with frontier as well. Thus,
$B(\vec x_1, r_1 / 2) \cap (A \cup Fr(A)) = \emptyset$.
But $\vec x_1$ was a point such that all balls around it intersected with $Cl(A)$. Thus, we reached a contradiction and $Cl(A)$ is a closed set. $\Box$

I feel like the part where I said any smaller ball around $\vec x_1$ would not even include a frontier point is a bit fishy. I thought that ball intersects with the frontier set but not the set $A$ itself. So it must be like tangent to the frontier, I thought. But how can I make this rigorous? And is this proof very amateurish?
 A: I'll denote the boundary ("frontier") by $\partial A$ instead.
There are always three options in the relation between $x$ and a set $A$.


*

*$\exists r>0 : B(x,r) \subseteq A$;these points are called in interior of $A$.

*$\exists r>0: B(x,r) \subseteq X\setminus A$ (or $B(x,r) \cap A = \emptyset$); these $x$ together are called the exterior of $A$, which is the interior of $X\setminus A$.

*The only remaining option: $\forall r>0 : B(x,r) \nsubseteq A \text{ and } B(x,r) \nsubseteq X\setminus A$. This is equivalent too $$\forall r>0: B(x,r) \cap A \neq \emptyset \text{ and } B(x,r) \cap (X\setminus A) \neq \emptyset$$ 
or $x \in \partial A$.


So $X$ is a disjoint union of $\operatorname{int}(A)$, $\operatorname{ext}(A) = \operatorname{int}(X\setminus A)$ and $\partial A$, for any $A \subseteq X$.
When $x \notin A \cup \partial A$, certainly $x \notin \operatorname{int}(A)$, so $x \in \operatorname{ext}(A)$, so we have $r>0$ such that $B(x,r) \cap A = \emptyset$. Now I use also that $B(x,r)$ is an open set, (for every $y \in B(x,r)$ there is some $r' >0$ (in fact $r' = r d(x,y) >0$ works) such that $B(y,r') \subseteq B(x,r)$ which is what the OP was going for (the idea at least)): If $y \in B(x,r) \cap \partial$, find $r' > 0$ as above tand $B(y,r')$ misses $A$ (because $B(x,r)$ does), so we cannot then have that $y \in \partial A$. So $B(x,r) \cap (A \cup \partial A)  =\emptyset$, showing that $A \cup \partial A$ is closed.
