Statement: Let $G$ be a subpresheaf of $F$. Then $G$ is closed if and only if for each $x \in FC$: if $R:= \{f: C' \to C \ | \ F(f)(x) \in G(C')\}$ covers $C$, then $x \in GC$.
I am having trouble with the direction where we assume $G$ is closed. It follows from the equivalence of Grothendieck topology - Lawvere-Tierney topology - universal closure operation that the $\bar{G}(C) = \{x \in FC \ | \ J_C(\phi_C(x)) = max(C)\}$ where $\phi$ classifies $G$. By the assumption, it suffices to prove that for any $x \in FC$, we have that $J_C(\phi_C(x)) = max(C)$.
Call the subobject classifier $1 \xrightarrow{t} \Omega$, then $t_C$ sends the unique element of $1(C)$ to the maximal sieve on $C$, so I think we can maybe pullback $R$ over some suitable morphism and then use that $$GC \to 1(C) \to 1(C) = GC \to FC \xrightarrow{\phi_C} \Omega_C \xrightarrow{J_C} \Omega_C$$
(by commutativity of the subobject classifier pullback)
I think the easiest way is chasing through the different ways to construct one kind of topology from another.
Recall that the classifying map $\phi$ of $G$, as a subobject of $F$, is given by: $$ \phi_C(x) := \{f: C' \to C \mid F(f)(x) \in G(C')\}. $$ If my guess about the notes you are using is correct, this should be around page 90, otherwise it is a quick exercise in writing out definitions. Note that this is precisely the sieve from the statement you want to prove.
Now, fix $x \in F(C)$ and suppose that $\phi_C(x)$ is covering, we will prove that $x \in G(C)$. In particular $Id_C^*(S_C(x))$, which is just $S_C(X)$, is covering. Recall that the Lawviere-Tierney topology associated with a Grothendieck toplogy is given by $$ J_C(S) = \{h: C' \to C \mid h^*(S) \text{ is covering}\}. $$ This means that $Id_C \in J_C(\phi_C(x))$. So $J_C(\phi_C(x)) = \max(C)$, the maximal sieve on $C$.
For the final bit, recall that the closure operation associated with a Lawvere-Tierney topology is given by: $$ \bar{G}(C) = \{x \in F(C) \mid J_C(\phi_C(x)) = \max(C)\}. $$ So we conclude from the above that $x \in \bar{G}(C) = G(C)$, where the last equality follows because $G$ is closed.
You did not ask about it, but let me include the proof of the converse for completeness.
Clearly $G(C) \subseteq \bar{G}(C)$ for all $C$, so we prove the inclusion in the other direction. Let $x \in \bar{G}(C)$, then $J_C(\phi_C(x)) = \max(C)$, so $Id_C \in J_C(\phi_C(x))$. Thus by definition of $J_C$, we have that $\phi_C(x) = Id_C^*(\phi_C(x))$ is covering. Then by the assumption from the statement, that means that $x \in G(C)$, as required.