Easy example:

If B is an infinite family of open sets whose union contains [0,1], then there is a finite subset of B which also has this property. Proof:

1. Let x be the lowest upper bound of all real numbers in the interval [0,1] such that there exists a finite subfamily of B that covers [0,x].

2. There is an open interval covering 0, so x is larger than 0.

3. Assume x!=1. We will derive a contradiction. Pick an interval in B which contains x. Then it contains a number a bit smaller than x, call it a. Then there is a finite subfamily covering [0,a], so a finite family covering [0,x]. Then the open interval containing x also extends a bit further than x, so x is not the upper bound.

QED