Actually it is not correct. The idea is right but not the details.
If is an open set then .
Now you can say that .
The other way is trivial.
So I did this:Prove that a subset of a metric space is open iff it is a union of open balls.
Suppose a subset of a metric space is open.
Let be a metric space.
Let
Then there exists some such that .
But is an arbitrary element of .
Therefore centre a ball around every point in .
.
(Since the centre of each of these balls is a point in , the union of them must be the entire set .)
Conversely, suppose a subset of a metric space is .
Then by definition is open.
Is this right?