Is it proof by contradiction? In a proof of the statement "A closed ball $\bar{B}\left(a,r\right)$ is a closed set", author first took complement of  $\bar{B}\left(a,r\right)$ and proved it an open ball.
 I was wondering whether the author used  proof by contradiction here.   
 A: No, it's not a proof by contradicition. He merely used the fact that $\overline B(a,r)$ is closed, if and only if,  $X\setminus \overline B(a,r)$ is open. (I suppose here $X=\Bbb R^n$, for some $n\in \Bbb N$).
He could have used proof by contradiction to prove that $X\setminus \overline B(a,r)$ is open or to prove facts leading up to the above theorem, but explicitly he didn't prove that $\overline B(a,r)$ is closed by contradiction.
A: I guess you mean the author took the complement of $\bar B(a,r)$ and proved it is an open set. This is a direct proof, since we usually define closed sets as those whose complement is an open set. Now, I guess you define open sets as those with the following property: given any point $x$ in the set, call it $O$, there is an open ball $B(x,r)$ about $x$ such that $x\in B(x,r)\subseteq O$.
The definition of open and closed may be slightly confusing at first. In everyday language, open means exactly not closed and closed means exactly not open. However, the definition in mathematics has a slight difference:

A set $A$ is open if and only if its complement (what is not $A$, in contrast to what is $A$) is closed.

In particular (we consider the usual metric $|x-y|=d(x,y)$ in $\Bbb R$)


*

*There are sets that are closed, but not open: $[a,b]$

*There are sets that are open, but not closed: $(a,b)$

*There are sets that are not closed, nor open: $[a,b)$

*There are sets that are both open and closed: $\Bbb R$

A: No, the author did not use proof by contradiction.
What is proof by contradiction? If you want to prove that a statement is true, you assume that the statement is false and derive a contradiction.
Note that the negation (the opposite) of being open is not being closed.
In your case, the author simply used that a set is closed if and only if the complement is open.
