Question

A set defined by a local property imply that it is an open set?

Answer 1

You should probably cite more of the proof, as is that can be a lot of things. "Based on a local property" is not very precise.

If you mean something like "The underlying space fulfills this property locally, i. e. for each point, we can find a neighbourhood where the property is fulfilled", then you can always choose the local subset (the neighbourhood) open, that follows from the definition of a neighbourhood ("contains an open set that contain the point in question").

Answer 2

Going out on a limb, I assume you have a topological (or metric, or normed) space X and the set in question is of the form S={x in X | x has property P}. And that P is a local property in the sense that if x has the property P, then so does every point in a suitable open neighborhood of x. Then this neighborhood would, by definition, be contained in S. And since this is true of every point of S, there is an open neighborhood around every point of S, which is also contained in S. Which makes S open.