It is well-known that the stalks of the skyscraper sheaf $\mathcal{G}_P$ (where $\mathcal{G}_P(U)=G$ if $P \in U$ and $\mathcal{G}_P=(0)$ otherwise). Indeed, stalks are of the form
$(\mathcal{G}_P)_q=G$ if $q \in \overline{\{p\}}$ and $(\mathcal{G}_P)_q=(0)$ otherwise. Let then $q \in \overline{\{p\}}$. Then any open neighborhood containing $q$, must also contain $p$, whence
$$ (\mathcal{G}_P)_q=\lim\limits_{\substack{\rightarrow \\ q \in V}} \mathcal{G}_P(V)=\lim\limits_{\substack{\rightarrow \\ q \in V}}=G $$
(the direct limit of copies of the same abelian group agrees with the group itself?)
Now, consider $q \notin \overline{\{p\}}$. Then $V=X \setminus \overline{\{p\}}$ is an open neighborhood of $q$ not containing $p$ and, hence, $\mathcal{G}_P(V)=\emptyset$. Thus, note that if we take any element of the stalk at $q$, i.e. $[(s,W)]_\simeq$, then $W \cap V$ does not contain $p$ and, thus, the section is the zero section. Is it my argument correct?