I am having difficulty trying to understand an innocuous statement in Hartshorne regarding stalks. The stalk at a point $p$ is defined to be the direct limit of $\mathscr{F}(U)$ where $U$ is an open neighborhood of $p$. I do understand that if $s \in \mathscr{F}(U)$ and $t \in \mathscr{F}(V)$ and there exists an open neighborhood $W$ of $p$ such that $W \subset U \cap V$ and $s |_W = t |_W$ then $\langle U, s \rangle = \langle V, t \rangle$. But why is the converse true?
My attempt: Let $\phi_O : \mathscr{F}(O) \to \mathscr{F}_p$ be the map such that $\phi_O(f) = \langle O, f \rangle$. Let $W$ be an open neighborhood of $p$ such that $W \subset U \cap V$. Then by definition of direct limit, we have $\phi_W(r_{U, W}) = \phi_U$ and $\phi_W(r_{V, W}) = \phi_V$. Now let $s \in \mathscr{F}(U)$ and $t \in \mathscr{F}(V)$ be such that $\langle U, s \rangle = \langle V, t \rangle$. Then it follows that
$$
\phi_W(s |_W) = \phi_W(r_{U, W}(s)) = \phi_U(s) = \langle U, s \rangle = \langle V, t \rangle = \phi_V(t) = \phi_W(r_{V, W}(t)) = \phi_W(t |_W).
$$
Hence, $s |_W - t |_W \in \ker \phi_W$. Since we want $s |_W = t |_W$, it would be ensured once we have $\ker \phi_W = 0$. But how can we choose $W$ to be such that this is satisfied.