Claim. Let $X$ be an integral scheme, and $\emptyset\neq V\subseteq U$ for some open subsets $V,U$ of $X$. Then $ \rho_{UV}:\mathcal{O}_X(U)\to\mathcal{O}_X(V)$ must be injective.
Here is my proof of the claim, which requires your criticism:
Proof. Let $s\in\mathcal{O}_X(U)$ be such that $s|_V:=\rho_{UV}(s)=0\in \mathcal{O}_X(V)$. Let $p\in U$ be an arbitrary point and $W$ be a non-empty open subset of $U$ containing $p$. Because $X$ is integral, hence irreducible, we must have $U$ an irreducible dense open subset of $X$. Moreover, $V$ is also a dense and irreducible open subset of $U$. Hence $V\cap W\neq \emptyset$, and $s_p=(s|_{V\cap W})_p=0$. Now if we let $p\in U$ vary, we would have $s_p=0$ for all $p\in U$, which means $s=0\in\mathcal{O}_X(U)$.
Issue. In the proof above, we cannot have $p\in V$, which will stop our further discussion about $(s|_{V\cap W})_p$.
A possible fix that needs your critism. Instead of considering points by points, we consider the vanishing points of $s$ over $U$, i.e. $$V(s):=\{\mathfrak{p}\in U| \ s_\mathfrak{p}\in \mathfrak{m}_\mathfrak{p}\subseteq \mathcal{O}_{X,\mathfrak{p}}\}$$ This set is closed in $U$ (4.3.F Vakil). And since for any $\mathfrak{p}\in V$, we have $s_\mathfrak{p}=(s|_V)_\mathfrak{p}=0\in \mathfrak{m}_\mathfrak{p}$, hence $V\subseteq V(s)$. Now taking the closure, we see that $U=V(s)$. If we cover $U$ with affine opens $U_i\cong \operatorname{Spec}A_i$, then for all $\mathfrak{p}\in U_i$, the fact that $s_\mathfrak{p}=0$ implies $s|_{U_i}\in \mathfrak{p}$. Hence $s|_{U_i}\in Nil(A_i)=\{0\}$ by reducedness. Now such $U_i$'s cover our $U$, which allow us to conclude that $s=0$.
Any help is appreciated! Thanks for reading.