This is exercise I-8 from Geometry of schemes by Eisenbud and Harris.
Let $\mathscr{F}$ be a sheaf, then topologize the union $\overline{\mathscr{F}}=\bigcup_{x \in X}\mathscr{F}_x$ by taking basic open set of this form $$\mathscr{V}(U,s)= \{(x,s_x): x \in U\}$$ Where $s \in \mathscr{F}(U)$ and $s_x$ is the corresponding equivalence class in the stalk.
It is clear that there is a natural continuous "projection" $\pi:\overline{\mathscr{F}}\longrightarrow X$
I have trouble proving the following:
Suppose $\sigma:U \longrightarrow \overline{\mathscr{F}}$ be a continuous map such that $\pi \circ \sigma$ is the identity on $U$.
Then there exists $s \in \mathscr{F}(U)$ such that $s_x = \sigma(x)$ for all $x \in X$.
Here is my idea, for every $x \in X$, it is clear that $\sigma(x) \in \mathscr{F}_x$. Then there exists open set $V_x$, $t^x \in \mathscr{F}(V_x)$ such that $\mathscr{V}(V_x,t^x)$ is a basic open neighbourhood containing $\sigma(x)$. It is not hard to see that for all $x' \in V$, $t^x_{x'} = \sigma(x') \in \mathscr{F}_{x'}$.
Then $\{V_x\}_{x \in U}$ is an open cover for $U$ and apply the sheaf axiom on the collection $\{t^x\}_{x \in X}$.
By I'm stuck on one little detail. Let $x,y \in U$ and suppose $V = V_x \cap V_y \neq \emptyset$. Why is it true that $t^x|_V = t^{y}|_V$?.
What I know is that for all $w \in V$, $t^x_{w} = t^{y}_{w} = \sigma(w) \in \mathscr{F}_w$. But that only means there exists $W \subseteq V$ containing $w$ such that $t^x|_W = t^{y}|_W$. This is a slightly weaker statement, so how should I proceed?