The following is exercise 6, chapter 0 from Johnstone's Topos Theory:
Suppose $F\rightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $\sigma\in G(U)$ for some open $U\subseteq X$. Prove that there is a unique largest open set $V\subseteq U$ such that $\sigma_{|V}$ is in the image of $F(V)$.
Unicity is easy, because given a family of such subsets $\{V_i\}_{i\in I}$ then the union $\bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:
$X=\{*\}$ the singleton space
$F\in Shv(X)$ the constant sheaf sending all opens to the set $\{a\}$
$G\in Shv(X)$ the constant sheaf sending all opens to $\{a,\sigma\}$
Then $F\rightarrowtail G$, but a $V$ as above does not exist for any choice of $U\subseteq X$ and taking $\sigma\in G(U)$.
If $A$ is a sheaf, then $A(\varnothing)$ is a one-point set.
So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = \{ a, \sigma \}$ and $G(\varnothing) = \{ * \}$.
Going back to the generic problem, $\sigma|_\varnothing$ is the unique element of $G(\varnothing)$, which is the image of the unique element of $F(\varnothing)$, so there's your existence proof.