A proof that the closure of a set A is equal to the union of A and its limit points Wanted to check if this proof is valid (not completely sure about the end):
We claim that $$\bar A = A \cup A'$$
Proof:
Take some a $\in A \cup A'.$ If $a \in A$ then $a \in \bar A$. If  $a \in A'$ then consider a sequence $(a_n) \rightarrow a$, where  $(a_n) \in A$  $\forall n \in \mathbb N $. Then since $\bar A$ is closed it contains all its limit points and so $a \in \bar A$. Thus $\bar A \supset A \cup A'$.
Now take some $x \in \bar A$. Obviously if  $x \in A \subset \bar A$ then $ x \in A \cup A'$. Consider some $y \in \bar A \setminus A$. We wish to show that $y \in A'$.
Assume otherwise, that is that $y \not\in A'$. Then $y$ is not a limit point of $A$ and so if we remove $y$ from $\bar A$ it is still closed. But this contradicts the minimality of $\bar A$ so $y$ is a limit point of $A$ and we have equality.
How do I rigorously show the contradiction here (or is it okay)?
 A: How do you conclude that removing a non-limit point from $\overline A$ leaves you with a closed set, without using the theorem that you are proving?
Better argue explicitly that there is an open ball $B$ around $y$ that avoids $A$, so $\overline A\setminus B $ is the intersection of two closed sets ($\overline A$ and the complement of the open set $B$) and therefore still a closed set (that contains $A$).
A: Yet another way to prove it:
Let $(X,d)$ be a metric space. If an adherent point of a set $A \subset X$ is defined as a point $x \in X$ such that for every $r > 0$ the open ball $B_{A}^{x}(r) = B^{x}(r) \cap A$ of center $x$ and radius $r$ is $\neq \varnothing$, if a limit point of $A$ is defined as a point $x \in X$ such that $B_{A}^{x}(r)\setminus \{ x\} \neq \varnothing$ for every $r > 0$, and if the closure $\overline{A}$ of $A$ is defined as the set of all adherent points of $A$,  then it is easy to see the statement. 
Indeed, it follows from the definition that every point of $A$ is an adherent point of $A$, so $A \subset \overline{A}$. But by definition every limit point of $A$ is an adherent point of $A$; hence $A' \subset \overline{A}$. Then we have $A \cup A' \subset \overline{A}$. Conversely, let $x \in \overline{A}$. Then by definition $x$ is an adherent point of $A$; suppose $x \notin A \cup A'$. Then $r := \inf_{y \in A}d(x,y)$ is $>0$, and hence $B_{A}^{x}(r/2) = \varnothing$, implying that $x$ is not an adherent point of $A$; this is a contradiction.
