If $X$ is countably compact, then since $\{X-C_i\}$ has no finite subcovering of $X$ , it does not cover $X$ . The other direction: suppose that there is a countable covering $\{U_n\}$ that has no finite subcovering, then $\{X-\cup_{i<n}U_i\}$ is a nested sequence of nonempty closed sets and it has a nonempty intersection. This contradicts the assumption that $\{U_n\}$ is a covering of $X$ . Note that the second part of the proof works for countable coverings only. For uncountable coverings there is no general way to order them in a sequence so that the sequence will contain all the sets in the collection and the complements of the partial unions of the sequence will generate a nested sequence of closed sets. Instead, we use the finite intersection property.