Let A be a non-empty subset of R (real numbers) and a an upper bound in R for A. Suppose that every open interval I containing a intersects A (so the intersection is non-empty). Show that a is a least upper bound for A.
The Attempt at a Solution
I've seen the prettier proof which involved a contradiction, but just wanted to know if my rather non-rigorous argument makes sense, and if it can be made more precise:
Let b=sup(A). Let an interval I=(x,y) satisfy the above (contains a, intersects A). If we take the limit as y approaches b, then a must approach b=sup(A) and so since the limit of this I is just another open interval, and a=sup(A) in this limit, then a=sup(A).