Exercise 1, chapter 3, Sheaves in Geometry and logic I'm working on the following exercise:

Let $X$ be a topological space. For a sieve $S$ on an open subset $U$
of $X$ define $S$ covers $U$ iff $U$ is the union of the sets in $S$.
Prove that this defines a Grothendieck topology on the partially
ordered set $Op(X)$ of all open subsets of $X$.

I've proved the maximal sieve axiom and the stability axiom, but I couldn't prove transitivity.
My attempt: In topological terms, transitivity is equivalent to "If $S$ covers $U$ and $R$ is a sieve on $U$ such that for every $h:V \subseteq U$,  $\{W \subseteq V / W \in R\}=h^{*}(R)$ covers $V$, then $R$ covers $U$". Clearly, $\bigcup R \subseteq U$. Now,
$$U= \bigcup S = \bigcup \{V / V \in S\} = \bigcup \{\bigcup\{ W \subseteq V / W \in R \} / V \in S \}$$
If $\bigcup\{ W \subseteq V / W \in R \}$ was in $R$ for every $V$, we would have the desired result; however, $R$ is closed under "subsets" not necessarily under unions.
Could you help me?
 A: In the comments Zhen Lin already points out an issue with your statement. The correct statement that you have to prove is below. The correct statement is actually a little bit stricter, because there is an extra condition on $h$.
In my notation I will identify an arrow $V \subseteq U$ with $V$. So a sieve $S$ on $U$ is just a set of subsets of $U$ and it is covering iff $\bigcup S = U$.

If $S$ covers $U$ and $R$ is a sieve on $U$ such that for every $V \in S$ we have that $V^*(R) = \{W \subseteq V : W \in R\}$ covers $V$ then $R$ covers $U$.

Proof 1. The direction $\bigcup R \subseteq U$ is automatic, so we prove the converse. Let $x \in U$ and let $V \in S$ be such that $x \in V$. Then $V^*(R)$ covers $V$, so there is $W \subseteq V$ with $x \in W \in R$. Hence $x \in \bigcup R$, and so we conclude that $U \subseteq \bigcup R$, as required.
Proof 2. Oneliner based on your proof, but maybe harder to read:
$$
U = \bigcup S = \bigcup \{V \subseteq U : V \in S\} = \bigcup \{ \bigcup V^*(R) : V \in S\} \subseteq \bigcup R.
$$
The last inclusion follows because $\bigcup V^*(R) \subseteq \bigcup R$ for any $V \subseteq U$.
