Open and closed cover for topological spaces I have the following question: 
Given a topological space $X$ and a open finite cover $\mathcal{U}=\{U_\lambda \mid \lambda \in I\}$ of $X$, what conditions (if any) do I need to ensure that one can find, for each $\lambda$, a closed set $K_\lambda\subset U_\lambda$, such that $\{\text{Int}(K_\lambda) \mid \lambda \in I\}$ also covers $X$. 
 A: You can certainly do it if $X$ is normal. Let $\mathscr{U}_0=\{U_1^{(0)},\ldots,U_n^{(0)}\}$ be a finite open cover of the normal space $X$. Let
$$F_1=X\setminus\bigcup_{k=2}^nU_k^{(0)}\;;$$
$F_1$ is a closed subset of $U_1^{(0)}$, so there is an open $U_1^{(1)}$ such that 
$$F_1\subseteq U_1^{(1)}\subseteq\operatorname{cl}U_1^{(1)}\subseteq U_1^{(0)}\;,$$ 
and we set $K_1=\operatorname{cl}U_1^{(1)}$. For $k=2,\ldots,n$ let $U_k^{(1)}=U_k^{(0)}$, and let 
$$\mathscr{U}_1=\left\{U_1^{(1)},\ldots,U_n^{(1)}\right\}\;;$$
$\mathscr{U}_1$ is still an open cover of $X$.
Given $\mathscr{U}_{k-1}$ for some $k\in\{1,\ldots,n\}$, let
$$F_k=X\setminus\bigcup\big(\mathscr{U}_{k-1}\setminus\{U_k^{(k-1)}\big)\;;$$
$F_k$ is a closed subset of $U_k^{(k-1)}$, so there is an open $U_k^{(k)}$ such that 
$$F_k\subseteq U_k^{(k)}\subseteq\operatorname{cl}U_k^{(k)}\subseteq U_k^{(k-1)}\;,$$ 
and we set $K_k=\operatorname{cl}U_k^{(k)}$. For $\ell\in\{1,\ldots,n\}\setminus\{k\}$ let $U_\ell^{(k)}=U_\ell^{(k-1)}$, and let 
$$\mathscr{U}_k=\left\{U_1^{(k)},\ldots,U_n^{(k)}\right\}\;$$
as before, $\mathscr{U}_k$ is still an open cover of $X$, and we can continue the recursive construction through to $\mathscr{U}_n$. Then for $k=1,\ldots,n$ we have $U_k^{(n)}\subseteq\operatorname{int}\operatorname{cl}U_k^{(n)}=\operatorname{int}K_k$, so $\{\operatorname{int}K_k:k=1,\ldots,n\}$ covers $X$, and $K_k\subseteq U_k^{(0)}$ for $k=1,\ldots,n$.
