Suppose $G$ is a subpresheaf of $F$. I want to prove: if $G$ is a sheaf, then $G$ is closed in $Sub(F)$.
My attempt so far: We want to prove that $\overline{G} = G$. It is sufficient to show for every $x \in F(C)$ that if $\{\alpha : D \to C \ | \ x \cdot \alpha \in G(D)\} = S \in Cov(S)$, then $x \in G(C)$. So suppose the former holds. Then we have a compatible family $\{x \cdot \alpha \in G(D)\}_{\alpha : D \to C \in S}$ for the cover $S$ of $C$. Hence, there exists a unique amalgamation $x' \in G(C)$ such that $x' \cdot \alpha = x \cdot \alpha$ for all $\alpha : D \to C \in S$. Now can we infer somehow that $x' = x$, am I on the right track at all?
Yes, $x$ is the amalgamation of its restrictions along the maps in any covering. This is the uniqueness part of the existence and uniqueness of amalgamations.
EDIT: The result is not true if $F$ is only assumed to be a presheaf. The presheaf $F$ on $a\to b$, with the trivial topology (every sieve is covering) sending $a$ to a singleton and $b$ to a 2-element set, admits two distinct subobjects $G$ which are sheaves but are not closed.