Consider $\omega+1$ with the interval topology, that is $U\subseteq (\omega+1)$ is open if and only if $U\subseteq\omega$ or $(\omega+1)\setminus U$ is finite.

We write $(\omega+1)^\omega$ for the collection of all functions $f:\omega\to(\omega+1)$. Let $\Box (\omega+1)^\omega$ be the set $(\omega+1)^\omega$ endowed with the [box topology][1], where each factor $(\omega+1)$ carries the interval topology.

For each $f\in(\omega+1)^\omega$ we consider the following clopen neighborhood $U_f$ of $f$: 

$U_f = \{g\in(\omega+1)^\omega:\forall n\in\omega\big((f(n) < \omega \Rightarrow g(n) = f(n)) \text{ and } (f(n) = \omega \Rightarrow g(n) \in [n,\omega])\big) \}$

Let ${\cal U} = \{U_f: f\in(\omega+1)^\omega\}$. Is there an open covering ${\cal V}$ of $(\omega+1)^\omega$ with the following properties?

 - each member of  ${\cal V}$ is contained in some member of ${\cal U}$;
 - for $V\neq W \in {\cal V}$ we have $V\cap W = \emptyset$.


**Note:** This is one special case of [this][2] question, see Joel David Hamkins' comment below.


  [1]: http://en.wikipedia.org/wiki/Box_topology
  [2]: http://%20mathoverflow.net/q/194281/1946