There is a neat proof of that result here.
A function is continuous on an open interval if it is continuous on every closed subinterval. So the proof given for closed intervals implies that the result is also true for open intervals.
But a convex function defined on an open interval may not have an extension to the closure of the interval (it could go off to infinity there). So I don't think that you can deduce the result for open intervals by taking the closure of the interval.