Let $X=Spec(A)$ be a noetherian affine scheme. Let $I_1, \ldots, I_n$ be ideals of $A$ such that $I_i + I_j = 1$ for all $i \neq j$.
Define $X_i = Spec(A/I_i)$ so that X is the disjoint union of the closed subschemes $X_i$.
Is something wrong with the following (and what if this is the case) : each $X_i$ is an open subscheme of $X$ since it is the complementary of a finite union of closed subscheme.
The condition that the ideals pairwise sum to $1$ gives that the sets $\mathrm{Spec}(A/I)$ are pairwise disjoint, but you have no condition on the ideals that guarantees that their union is all of $\mathrm{Spec}(A)$. For example, your condition does not preclude me from just removing one of the ideals to get a different collection that satisfies the same condition.
The condition that the union is $\mathrm{Spec}(A)$ is that the intersection $\bigcap I$ of all the ideals is contained in the nilradical of $A$, or equivalently that the product $\prod I$ is contained in the nilradical (I'm assuming your collection of ideals is finite, I don't think these conditions are equivalent otherwise). If by union you mean scheme theoretic union and not just union of the underlying topological spaces then actually I believe you want $\bigcap I = 0$.