Theorem 3.42 (Wedhorn and Görtz "Algebraic Geometry") states that, for $X = \operatorname{Spec} A$, there is a bijection between ideals of $A$ and closed subsechemes of $A$ given by $\mathfrak{a} \mapsto V(\mathfrak{a})$ (where the structural sheaf in $V(\mathfrak{a})$ is given by the pushforward of the structural sheaf of $\operatorname{Spec}(A/\mathfrak{a})$ via the canonical homeomorphism $\operatorname{Spec} A/\mathfrak{a} \to V(\mathfrak{a})$).
In order to prove this, he argues that it is enough that prove that if $i: Z \to X$ is the inclusion morphism and the associated map $\varphi: A \to \Gamma(Z, \mathscr{O}_Z)$ is injective, then $i$ is an isomorphism.
First, he shows that from the previous hypothesis, it follows that $i$ is a homeomorphism. So now we assume that $Z$ and $X$ have the same underlying topological space. In order to show that $\mathscr{O}_X \to \mathscr{O}_Z$ is an isomorphism, we have only to show that it is injective (because $i$ is a closed immersion). Then they set of to prove injectivity in the stalks. Let $x \in Z$, with $\mathfrak{p}_x$ the associated prime ideal in $A$, and consider the morphism $\mathscr{O}_{X, x} = A_{\mathfrak{p}_x} \to \mathscr{O}_{Z, x}$. It is enough to show that for each $g \in A$ such that $\frac{g}{1} \in \operatorname{ker}(\mathscr{O}_{X, x} \to \mathscr{O}_{Z, x})$, $g = 0$. To prove this, the book says we whould consider a finite open covering $Z = U \cup \bigcup_i U_i$ satisfying:
$(U, \mathscr{O}_Z|_U)$, $(U_i, \mathscr{O}_Z|_{U_i})$ are affine schemes.
$\varphi(g)|_U = 0, x \in U$ and $x \notin U_i$ for each $i$.
The first time I wrote the question I assumed that $U$ was fixed because we get it by choosing an affine neighborhood of $x$ such that $\varphi(g)|_U = 0$. But I guess it must be defined at the same time as the other $U_i$. I don't know why this open cover exists.