If a topological space $X$ has an open covering $\{ U_i \}$ such that every $U_i$ is irreducible and that for each $i,j$, $U_i \cap U_j \neq \emptyset$, then is $X$ irreducible?
I found Space admitting an irreducible connected open covering is irreducible , but I think this is wrong.: If this answer were correct, every scheme would be affine.
It is true. Let $U$, $V$ be any two non-empty open subsets of $X$. We want to show that $U\cap V\ne \emptyset$. Since $\{U_i\}$ is a covering, there must be some $i,j$ such that $U_i\cap U\ne\emptyset$ and $U_j\cap V\ne\emptyset$. Since $U_i\cap U$ and $U_i\cap U_j$ are non-empty open subsets of $U_i$ and $U_i$ is irreducible, $U_i\cap U_j\cap U=(U_i\cap U_j)\cap(U_i\cap U)\ne\emptyset$. Finally, since $U_i\cap U_j\cap U$ and $U_j\cap V$ are non-empty open subset of $U_j$ and $U_j$ is irreducible, then $U_i\cap U_j\cap V\cap U=(U_i\cap U_j\cap U)\cap(U_j\cap V)\ne\emptyset$.
I can't quite weight on the scheme part, but probably what you have for those is a situation where $U_1\cap U_2$ and $U_2\cap U_3$ are non-empty, while $U_1\cap U_3=\emptyset$.
Added: For the record, let's recall the definition of irreducible (or hyperconnected) topological space: encyclopediaofmath.org, Wikipedia.