Let $X$ be a topological space. A presheaf $G$ of abelian groups on $X$ is a sheaf if for every open $V\subset X$ and every covering $\cup_{\alpha\in A} V_\alpha$ of $V$ the diagram $$ G(V) \to \prod_{\alpha\in A} G(V_\alpha) \rightrightarrows \prod_{(\alpha,\beta)\in A²} G(V_\alpha\cap V_\beta) $$ is an equalizer.
If the space $X$ is noetherian, a presheaf $G$ is already a sheaf if the sheaf condition holds only for finite coverings, i.e. for those coverings with $|A|<\infty$.
I wonder if this can be weakened even more to coverings by just two open sets $V_1\cup V_2=V$. Is this true? If yes, is there a reference in the literature?
This is almost true: in addition to requiring the gluing condition for covers by two open sets, you also need to require it for covers by zero open sets. I don't know a reference, but the proof is very easy: just use induction on the number of sets in the open cover. If $V=V_1\cup V_2\cup\dots V_n$ for $n>2$, let $U_1=V_1$ and $U_2=V_2\cup \dots\cup V_n$. By the induction hypothesis you can glue compatible sections on the $V_i$ uniquely to get a section on $U_2$, and then you can uniquely glue your section on $U_2$ with your section on $U_1$ to get a section on all of $V$. (The sections on $U_1$ and $U_2$ agree on $U_1\cap U_2$ by the uniqueness part of the induction hypothesis, since $U_1\cap U_2$ is covered by the $n-1$ open subsets $V_1\cap V_2,V_1\cap V_3,\dots,V_1\cap V_n$.)