Let $P: (OP_{X}) \to (Sets)$ be a sheaf, where $(OP_X)$ is just the category of open sets of a topological space $X$. Define $P_x$ to be the stalk of $x \in X$. Then we can define the Etale space to be $Et(P):=\amalg_{x \in X} P_x$, and the topology is given by a basis $\{U_s| U \ open\ in X\}$, where $U_s = \{S_x|x \in U\}$($S_x$ is the germ of $S$ at $x$). There is a natural projection $\pi: Et(p) \to X$, which is continuous and surjective.
Now the claim is that there is a natural continuous map $g: X \to Et(p)$ such that $\pi \circ g = id_{X}$.
My approach is to associate with each point $x \in X$ a section $s \in P(X)$, since $X$ is a natural open set containing $x$. However this choice of section seems to be too arbitrary to be continuous.
The reason that I ask about this is that the existence of a global section ensures that we can define $P^{+}$ the sheaf associated to $P$ because $P^{+}(U)$ is defined to be the set of all continuous sections on $U$ and the restriction of a global section onto each open subset ensures that $F^{+}(U)$ is not empty.
This answer gives a more precise account of this construction
Update: Inspired by the given link, given $s \in P(X)$ we can consider directly the map $g:X \to Et(P)$, $g(x) = s_x$. Now I want to show that it is continuous. $g(x) \in U_{s'}$ iff there is an open set $U'$ containing $x$ such that $r^{X}_{U'}s = r^{U}_{U'}s'$. What then?
