[proofplan] Fix a point $x \in X$. Local finiteness of the family $\mathcal U$ gives an open neighbourhood $W$ of $x$ that meets only finitely many members of $\mathcal U$. Every function indexed by a member of $\mathcal U$ disjoint from $W$ vanishes on all of $W$, because it is assumed to vanish outside its indexing [open set](/page/Open%20Set). Hence the only functions that can be nonzero somewhere on $W$ are indexed by that finite subfamily. [/proofplan] [step:Choose a neighbourhood meeting only finitely many cover elements] Fix $x \in X$. Since $\mathcal U$ is locally finite, there exists an open neighbourhood $W \in \tau$ of $x$ such that the indexing set \begin{align*} \mathcal F := \{U \in \mathcal U : W \cap U \neq \varnothing\} \end{align*} is finite. [guided] We begin at the point where local finiteness is used. Fix $x \in X$. The family $\mathcal U$ is locally finite, meaning that every point of $X$ has an open neighbourhood that intersects only finitely many sets in the family. Applying this definition at the chosen point $x$, we obtain an open neighbourhood $W \in \tau$ of $x$ for which the collection \begin{align*} \mathcal F := \{U \in \mathcal U : W \cap U \neq \varnothing\} \end{align*} is finite. This finite set $\mathcal F$ records precisely the possible indices whose supports can interact with $W$. The rest of the proof shows that if an index $U$ is not in $\mathcal F$, then the corresponding function $\phi_U$ has no room to be nonzero on $W$. [/guided] [/step] [step:Show all functions indexed outside the finite subfamily vanish on the neighbourhood] Let $V \in \mathcal U \setminus \mathcal F$. By the definition of $\mathcal F$, we have $W \cap V = \varnothing$, so $W \subset X \setminus V$. The support condition in the theorem statement gives $\phi_V(y)=0$ for every $y \in X \setminus V$. Therefore $\phi_V(y)=0$ for every $y \in W$. [/step] [step:Conclude that only finitely many functions can be nonzero on the neighbourhood] Define \begin{align*} \mathcal G := \{U \in \mathcal U : \text{there exists } y \in W \text{ with } \phi_U(y) \neq 0\}. \end{align*} The previous step proves that no element of $\mathcal U \setminus \mathcal F$ belongs to $\mathcal G$, so $\mathcal G \subset \mathcal F$. Since $\mathcal F$ is finite, $\mathcal G$ is finite. Thus $W$ is an open neighbourhood of $x$ on which only finitely many of the functions $\phi_U$ are nonzero somewhere. Since $x \in X$ was arbitrary, the conclusion holds for every point of $X$. [/step]