In my notes there is the following result:
Suppose $X$ can be covered by $r$-many open and path-connected sets $X=X_1\cup\ldots \cup X_r$. Assume that $H^\ast(X_i)=0$ for all $i=1,\ldots, r$. Then for any $\alpha_1,\ldots,\alpha_r\in H^{\ast\geq 1}(X)$, we have $\alpha_1\cup\ldots\cup\alpha_r=0$.
First question: When they write $H^\ast(X_i)=0$ they mean that the reduced cohomology $\tilde{H}^\ast(X_i)=0$ vanishes, correct? Otherwise this result would be boring, as then $X_i=\emptyset$, i.e. $X=\emptyset$. In other words, the condition is equivalent to the fact that $X$ can be covered by $r$-many contractible open sets, right?
Second question: What are some noteable applications of this result? Any cohomology rings that can be easily calculated using this proposition?
Notice that the cup product defines a map $$H^*(X,A) \times H^*(X,B) \longmapsto H^*(X,A\cup B).$$ If $U$ is a set in your cover (you need only assume that the inclusion $U\to X$ is nullhomotopic, not that $U$ is itself contractible), then $H^*(X,U)\to H^*(X)$ is onto (look at the long exact sequence), so that whenever you pick $r$ different cocycles, you can assume they come from relative cohomology, and then their product maps to $H^*(X,X)$, the zero group.
This fact relates the cup length of a space to its Lusternik–Schnirlemann category, namely it gives that $\mathrm{cl}(X) \leqslant \mathrm{cat}(X)$.
For example, you can cover $\mathbb CP^n$ by $n+1$ affine subsets, but not $n$, because the degree $2$ generator $\xi$ satisfies $\xi^n\neq 0$, so the cup length is at least $n+1$, and so $\mathrm{cat}(\mathbb CP^n) = n+1$.