This is question 2.5 of Qing Liu.

I am new in algebraic geometry and really stuck on it and can't do anything to solve it.

The question:
Let $F$ be a sheaf on $X$. Let $\operatorname{Supp} F=\{x\in X:F\_x\neq 0\}$. We want to show that in general, $\operatorname{Supp} F$ is not a closed subset of $X$. Let us fix a sheaf $G$ on $X$ and a closed point $x\_0\in X$. Let us define a pre-sheaf $F$ by $F(U)=G(U)$ if $x\_0\notin U$ and $F(U)= \{s\in G(U):s\_{x\_0}=0\}$ otherwise. Show that $F$ is a sheaf and that $\operatorname{Supp} F = \operatorname{Supp}G\setminus \{x\_0\}$.

I don't know how to solve this question:
To show a pre-sheaf is a sheaf I need to check the "uniqueness" and "gluing local sections".

For the uniqueness: Let $U$ be an open subset of $X$ , $s\in F(U)$, if $x\_0\notin U$ , then since $G$ is a sheaf, I don't see a problem for $F$ to be a sheaf.

If $s\in F(U)$ and $x\_0 \in U$ and $\{U\_i\}\_i$ be an open covering of $U$,then there exists an $i\_0$ such that $x\_0\in U\_{i\_0}$. the image of $s$ in the stalk $F\_{x\_0}$ is $s\_{x\_0}$. $F(U\_{i\_0})=\{s\in G(U):s\_{x\_0}=0\}$ by definition. I don't know what to do now? (so sorry and I know this is an easy question...)