The following exercise appears in the algebraic geometry textbook by Görtz and Wedhorn (Exercise 2.13 (b)):
Let $X$ be a connected topological space and assume that there exists a sheaf $\mathcal{F}$ such that $\mathcal{F} (X) \to \mathcal{F} (U)$ is bijective for all non-empty open sets $U \subseteq X$ and such that $\mathcal{F} (X)$ contains more than one element. Show that $X$ is irreducible.
Is the assumption that $X$ be connected necessary here?
Here's my argument that does not use connectedness.
Note that the assumption that all the restrictions $\mathcal{F} (X) \to \mathcal{F} (U)$ are bijective for $U \ne \emptyset$ implies that $\mathcal{F} (U) \to \mathcal{F} (V)$ is bijective for any pair of nonempty open subsets $V \subseteq U$. Indeed, in this case the restriction $\operatorname{res}^U_V\colon \mathcal{F} (U) \to \mathcal{F} (V)$ is given by $$\mathcal{F} (U) \xrightarrow{(\operatorname{res}^X_U)^{-1}} \mathcal{F} (X) \xrightarrow{\operatorname{res}^X_V} \mathcal{F} (V)$$
To show that $X$ is irreducible, we may show that any two nonempty open subsets $U,V$ have a nonempty intersection. For the sake of contradiction, assume that $U, V \ne \emptyset$ and $U\cap V = \emptyset$. Then the sheaf axioms for $\mathcal{F}$ imply that $$\mathcal{F} (U\cup V) \xrightarrow{\cong} \mathcal{F} (U) \times \mathcal{F} (V)$$ is a bijection, given by $s \mapsto (s|_U, s|_V)$. However, the restrictions $\rho_U\colon s \mapsto s|_U$ and $\rho_V\colon s \mapsto s|_V$ are bijective by our asumption on $\mathcal{F}$, and this would imply that the diagonal map $$\Delta\colon \mathcal{F} (U\cup V) \xrightarrow{(\rho_U,\rho_V)} \mathcal{F} (U) \times \mathcal{F} (V) \xrightarrow{\rho_U^{-1}\times \rho_V^{-1}} \mathcal{F} (U\cup V) \times \mathcal{F} (U\cup V)$$ is bijective. But as $\mathcal{F} (U\cup V) \cong \mathcal{F} (X)$ has at least two different elements $s \ne t$ by our assumption, the element $(s,t)$ is clearly not in the image of $\Delta$.
This contradiction shows that $U\cap V \ne \emptyset$.
Does this look right? Note that the fact that $X$ is connected was not used. (But I don't see how a disconnected $X$ can admit such a sheaf $\mathcal{F}$.)