Unfortunately this may not satisfy the base criterion. Imagine a set that looks like the capital letter V with a dot below. Given x and y, two points in V, x < y if x is lower than y, on the same branch of V. Call the lower dot q, and let q be less than all the points in V. Use q and the two points at the top of V to define two chains, two base open sets, whose only intersection is p, at the base of V. There is no chain, no base set, that contains only p. This violates the base criterion; we don't have a valid topology. In general, you will need to verify the base criterion when you establish an order topology.
If a set is linearly ordered, the order topology is valid. The intersection of two chains is bounded below by the larger of the two lower bounds, and above by the smaller of the two upper bounds, hence the intersection is always another base set.
If p is any point in a linear ordering, and the ordering has no least or greatest element, the union of chains with p as upper bound is open, as is the union of chains with p as lower bound. Thus p is closed. All points in such a topology are closed.
If the space includes a concept of distance, the ball topology has the set of open balls as its base. An open ball is the set of points < r distance away from a point p, for some real radius r and a point p in the space. This does not always satisfy the base criterion. However, a few constraints on the distance function make this a proper base. This is discussed in metric spaces.