This has been done before on this site, but I a wondering if there is another approach.
Let $F:\mathcal{F}\rightarrow \mathcal{G}$ be a monomorphism of sheaves of abelian groups. Take the natural morphism $\iota:\ker F\rightarrow \mathcal{F}$, and note that $F\circ \iota=0$ by definition. Since $F$ is mono this implies that $\iota$ is the trivial morphism, such that $\iota_U=0$ for all $U$. However on $U$, $\iota_U$ is the inclusion map by definition, so we have that for all $s\in \ker F_U$ $\iota_U(s) =s=0$, so $\ker F_U=0$ for all $U$. It follows that $\ker F$ is the trivial sheaf, so $(\ker F)_x$ is the trivial stalk. But $(\ker F)_x=\ker F_x$, hence $F_x$ is injective as desired.
Does this not work for some reason?