Proof: Since is assumed to be bounded we have . Bisect the closed interval into two intervals and . Let . Take some point . Bisect into two new intervals, and label the rightmost interval . Since there are infinite points in we can pick some , and continue this process by picking some . We show that the sequence is convergent. Consider the chain
By the Nested Interval Property we know that there is some contained in each interval. We claim . Let be arbitrary. The length for each is, by construction, which converges to . Choose such that for each that . Since and are contained in each interval, it follows that .