In the proof of the Theorem 36.1 the first step was to show the result of this question for finite families of open set covering $X$ . The essential part of the proof is to construct each $V_i$ such that $X-\cup_{j\neq i}V_j\subseteq V_i\subseteq\overline{V}_i\subseteq U_i$ . We need the first inclusion to ensure that it covers $X$ , and the last one for later construction of a partition of unity. In the finite case we construct $V_i$ ’s one-by-one such that $V_i$ contains $X-\cup_{j<i}V_j\cup_{j>i}U_j$ . Note that at the last step $V_n$ covers all of $X-\cup_{j<n}V_j$ , therefore, all points of $X$ are covered. If the covering is countable, and we use the same way to construct sets $V_i$ ’s, then, in general, we cannot guarantee that each point will be contained in some set $V_i$ . However, if every point $x$ is contained in a finite number of sets $U_i$ , then for some $n$ , $U_{n+1}\cup U_{n+2}\cup . . .$ does not contain $x$ and, since $V_n$ contains $X-V_1\cup V_2\cup . . .\cup V_{n-1}\cup U_{n+1}\cup U_{n+2}\cup . . .$ , $x$ must be contained in $V_1\cup . . .\cup V_n$ . Therefore, the collection of $V_i$ ’s constructed the same way as in the theorem, does cover all of $X$ .[ To construct a counterexample we would need a space such that it is a) normal, b) there is a countable cover by open subsets $U_n$ (not point-finite) such that there is no other cover $V_n$ such that $\overline{V}_n\subseteq U_n$ . I have been thinking about the example for a while, but have not figured out one yet. The fact that the space is normal implies, in particular, that the construction above is possible. On the other hand, it seems that the construction is, in some sence, necessary... What I mean is that for any set $V_n$ , a) $V_i$ must cover $X-\cup_{j\neq i} V_j$ , therefore, it must cover $X-\cup_{j<i}V_j\cup_{j>i}U_j$ , and b) it must be such that $\overline{V}_j\subseteq U_j$ . So, if there are $V_i$ ’s for $U_i$ ’s, then they can be constructed using the procedure above. Our only hope is to find a space such that whatever sequence $V_n$ we take using the procedure above, there will be a point such that it is not covered. In particular, when we define $V_1$ such that $X-\cup_{j>1}U_j\subseteq V_1\subseteq\overline{V}_1\subseteq U_1$ there must be some points in $U_1$ not covered by $V_1$ such that each is contained in an infinite number of sets $U_j$ . ]