Let $(\mathcal{F}_i)_{i \in I}$ be a family of sheaves (of sets) on a topological space $X$. I wanna find a counter-example to the assertion that $$U \mapsto \coprod_{i \in I} \mathcal{F_i}(U)$$ defines a sheaf $\mathcal{F}$ of sets on $X$, where $\coprod$ is the coproduct (i.e. the disjoint union) in Set.
I've already checked that $\mathcal{F}$ is a separated presheaf so the thing that has to go wrong is the glueability axiom but I'm having trouble coming up with a counter-example.
Take $X = [0, 1] \sqcup [2, 3].$ Suppose $\mathcal{F}_1, \mathcal{F}_2$ are two sheaves. Let $\mathcal{G}$ denote the disjoint union presheaf.
Then take some $s \in \mathcal{F}_1([0, 1])$ and some $t \in \mathcal{F}_2([2, 3]).$ We should be able to glue these to get an element of $\mathcal{G}([0, 1]\sqcup [2, 3]).$ But what element would it be? :)