Let $F$ be a presheaf over $X$. I'm trying to prove that if $F$ satisfies the first sheaf axiom (locality in the wikipedia terminology), then it satisfies the gluing axiom iff the morphisms $\varepsilon_U: F(U) \rightarrow \hat{F}(U)$ defined by $\varepsilon_U (f)(x):= f_x$ ($f_x$ being the stalk of $f$ at $x$) are surjective for all $U$ open in $X$.
For context, the definition of the sheafification I'm working with $\hat{F}(U) := \left\{ \alpha: U \rightarrow Et_F, \alpha \textrm{ continuous and } \alpha(x) \in F_x \forall x \right\}$, where the topology on $Et_F$ is such that the projection to $X$ as well as all functions of the form $[x \mapsto f_x]$ where $f \in F(U)$ are continuous.
I am basically having trouble using the definition of the topology on $Et_F$ to rigorously prove the claim. I argued as follows: Let $\alpha \in \hat{F}(U)$. Then $\exists f \in F(U), \varepsilon_U(f)=\alpha$ iff $f_x = \alpha(x)$ for any $x$ iff $\forall x \in U: \exists W_x$ a nbhd. of $x$ such that $f|_{W_x} = g^{\alpha, x}$, where $g^{\alpha,x} \in F(W_x)$ and implicitly then $g^{\alpha, x}_x = \alpha(x) = f_x$ at the level of stalks. This is almost what I want, except I don't know how to show that $g^{\alpha, x}$ agree on intersections (this would show that if $F$ has the gluing axiom, then indeed the wanted $f$ exists, it is just the result of glueing all $g^{\alpha,x}$). Conversely, I would have to show that any local $g \in F(W_x)$ are of the form $g^{\alpha,x}$ for some (same) $\alpha \in F(U)$. This, I think, is where I should use the topology on $Et_F$, but I can't figure out the technical details.
Suppose that for a covering $\{U_i\}$ you have a set of compatible elements $f^i \in F(U_i).$ You wish to show that they come from one global element $f \in F(X).$ To that end it is enough to construct a global section $s \in \hat{F}(X)$ such that $s$ is $x \mapsto f^i_x$ on each $U_i$. It will define a unique $f$ such that $f_x = f^i_x$ for $x \in U_i,$ so $f$ and $f^i$ will coincide in a vicinity of each $x$, thus $f_{U_i}=f^i$.
So, for $x \in U_i$ let's just set $s(x) := f^i_x.$ This is what we look for, as long as $s$ is continuous. But the topology on Et is defined so that taking germs is a continuous map, which you wrote at the end of the second paragraph.