Why is the etale map an open map? In Proposition 5.59 (page 276) of his book An Introduction to Homological Algebra, Rotman states that an etale map is always an open map on sheaf space. (5.59iii)

Proposition 5.59
  Let $\mathcal{S} = (E, p, X)$ and $\mathcal{S}' = (E', p', X')$ be etale-sheaves over a topological space $X$.  
(iii)
  Every etale-map $\varphi \colon \mathcal{S} \to \mathcal{S}'$ is an open map $E \to E'$.
Proof. The sheets form a base of open sets for $E$.

Why does this follow?
 A: Fact: If $f: X \to Y$ is a function between spaces and $\mathcal{B}$ is a base for the topology of $X$, then $f$ is an open map iff $$\forall B \in \mathcal{B}: f[B] \text{ is open in } Y$$
This is proved simply by observing that for families $A_i \subseteq X, i \in I$ we have that $f[\bigcup_i A_i] = \bigcup_i f[A_i]$ and knowing that open sets are unions of basic open sets and open sets are closed under all unions.
So you're done showing openness if the image of a "sheet" under an étale map is open.
This is clear, as the definition of étale-map implies that for a sheet $S$ of $p$, we have that $\varphi[S] = (p')^{-1}[p[S]]$, where the left to right inclusion follows from $p'\varphi=p$ and the right to left one uses the local injectivity of $p$. So $\varphi[S]$ is open as $p[S]$ is open by the local homeomorphism part of the protosheaf definition and $p'$ is continuous.
A: I don't think the hint given in the book works. But we can proceed as follow.
Lemma. Let $g:X \to Y$ be a continuous map.
Suppose that $f:Y \to Z$ and $h:X \to Z$ are local homeomorphisms, such that $fg=h$. Then $g$ is an open map.
Proof. Let $U$ be an open set of $X$. We want to prove $g(U)$ open.
For all $x \in U$, there exists open neighbourhoods $x \in U_1 \subset X$, $h(x)\in V \subset Z$ and $g(x)\in W\subset Y$, such that $h|_{U_1} : U_
1 \to V$ and $f|_W : W\to V$ are homeomorphisms.
$g$ continuous $\implies$ $g^{-1}(W)$ is an open set of $Y$ containing $x$. Put $U_0 = U \cap g^{-1}(W) \cap U_1$, then $U_0$ is an open neighbourhood of $x$. We have $f|_W \circ g(U_0) = h|_{U_1}(U_0) \implies g(U_0) = (f|_W)^{-1}(h|_{U_1}(U_0))$. Then $g(U_0)$ which is a subset of $g(U)$ containing $g(x)$ is open in $W$, thus open in $Y$. So $g(U)$ is open. $\square$
We apply the above lemma to $p' \varphi =p$, where $p'$ and $p$ are local homeomorphisms.
