On the page 178 of Dungundji's Topology,
Problem 6 of Section 1 says:
A grating of a space $Y$ is a family $\{V_{\alpha}\;|\;\alpha \in A\}$ of disjoint open sets such that $\{\overline{V_{\alpha}}\;|\;\alpha \in A\}$ is a covering of $Y$ and $\;Int(\overline{V_{\alpha}}) = V_{\alpha}\;$ for each $\alpha \in A.\;$ Let $\{V_{\alpha}\;|\;\alpha \in A\}$ and $\{W_{\beta}\;|\;\beta \in B\}$ be two gratings of $Y$. Prove: $$\{V_{\alpha}\cap W_{\beta}\;|\;(\alpha, \beta) \in A\times B\}$$ is a grating of $Y$.
So far I've succeeded in proving $$Int(\overline{V_{\alpha}\cap W_{\beta}}) = V_{\alpha}\cap W_{\beta} \quad\forall \alpha, \beta$$, but couldn't prove $\{\overline{V_{\alpha}\cap W_{\beta}}\;|\;(\alpha, \beta) \in A\times B\}$ covers $Y$ using given conditions only.
My question is: Is the statement of the problem right? I think $\{(-\infty,0), (1, \infty)\}\cup\{(\frac{1}{n+1},\frac{1}{n})\;|\;n \in \mathbf{N}\}$ and $\{(-\infty,-1), (0, \infty)\}\cup\{(-\frac{1}{n},-\frac{1}{n+1})\;|\;n \in \mathbf{N}\}$ are both gratings of $\mathbf{R}$, but their intersection
$$\{(-\infty,-1), (1,\infty)\}\bigcup\left\{(-\frac{1}{n},-\frac{1}{n+1})\;\big|\;n \in \mathbf{N}\right\}\bigcup\left\{(\frac{1}{n+1},\frac{1}{n})\;\big|\;n \in \mathbf{N}\right\}$$ is not a grating since $0$ does not belong to any of their closures. Am I thinking right? Or is there a way to prove coveredness of $Y$?
You are right.
This would be true if $(V_\alpha)_\alpha$ and $(W_\beta)_\beta$ had the property that for any $\alpha,\beta$, we have $\overline{V_\alpha\cap W_\beta}=\overline{V_\alpha}\cap \overline{W_\beta}$, but in your example, it is clearly not the case, as $(-1,0)\cap (0,1)=\emptyset$, but their closures intersect.