Closure of $(a,b)$ is [$a,b$] I'm trying to write more formal proofs for rather obvious/intuitive things. I wanted to know whether my proof for this question is good enough.


  Prove that the closure of the set $(a,b)$ is the set [$a,b$].


From a well known property we know that $Cl(A)=A \cup \operatorname{acc}(A)$

Then it is enough to prove that $a$ and $b$ are accumulation points of $(a,b)$.

I will be using the following definition for accumulation point : 


  $x$ is an accumulation point of the set A if $B(x,r)\cap(A \setminus \{x\}) \neq \varnothing \ \forall r&gt;0$.


Now to check whether $a$ and $b$ are accumulation points we need to check if there exists $r&gt;0$ such that $B(a,r)\cap ((a,b) \setminus \{a\}) = \varnothing$. 

$B(a,r)=(a-r,a+r)$

It is obvious that we can always find a  point such that belongs in the second half of the interval that is $x \in (a,a+r)$ and also belongs in $(a,b) \setminus \{a\}$.

So indeed $a \in \operatorname{acc}((a,b))$. 

Similarly $B(b,r)=(b-r,b+r)$ and there exists point $x$ that belongs in the first half of the interval and also belongs in $(a,b) \setminus \{b\}$

So these two points are indeed accumulation points. It follows that $Cl((a,b))=
[a,b]$

Is this solution correct? Also is there perhaps a simpler approach ?

EDIT: After reading through the answers I realized that we also need to prove that there are no other accumulation points outside the interval $(a,b)$. This can be done by picking suitable radius. There were couple of them suggested in comments/answers.
The solution is correct and very simple as far as it goes. 

However, it is not complete. You also have to prove that no point in $\mathbb R - [a,b] = (-\infty,a) \cup (b,\infty)$ is an accumulation point.
Put $$a_n=a+\frac{1}{n}$$
and
$$b_n=b-\frac{1}{n}$$

for large enough $n$,

$$(a_n,b_n)\in (a,b)^2,$$

and
$$\lim_{n\to+\infty}a_n=a$$
$$\lim_{n\to+\infty}b_n=b.$$

on the other hand, $(a_n)$ and $(b_n)$ are not stationnary sequences, thus $a$ and $b$ are accumulation points.