Suppose $X$ is a CW complex and $X_n$ is the $n$th skeleton. Suppose that for $k=0, \dots n$ we have defined partitions of unity $(\psi_\alpha^k)$ for $X_k$ subordinate to $(U_\alpha^k)$ satisfying the following property for each $\alpha \in A$ and each $k$: If $\psi_\alpha^{k-1} \equiv 0$ on an open subset $V \subset X_{k-1}$, then there is an open subset $V' \subset X_k$ containing $V$ on which $\psi_\alpha^k \equiv 0.$
This property is used in showing local finiteness of the partition of unity. If $x \in X_n,$ because $\psi_\alpha^n \equiv 0$ except when $\alpha$ is one of finitely many indices, and then the above property shows that $\psi_\alpha^{n+1} \equiv 0$ on $V'$ except when $\alpha$ is one of the same indices.
However, I cannot see how this is true. If the index set $A$ is infinite, then we cannot take intersections of the open sets. How can we ensure this?
The supports of the $\psi^{k-1}_{\alpha}$ form a locally finite cover of $X_{k-1}$ so for each $x\in X_{k-1}$ there is a nbhd $W$ of $x$ such that $W$ intersects only finitely many $\text{supp}\ \psi^{k-1}_{\alpha}$, say $\text{supp}\ \psi^{k-1}_{\alpha_{1}},\cdots,\text{supp}\ \psi^{k-1}_{\alpha_{1}}.$ For any index $\alpha\ \textit{not}$ among the $\{\alpha_i:1\le i\le n\},\ \psi_{\alpha}(W)=0.$ Then, by hypothesis, there is an open set $O\in \tau_{X_k}$ on which $\psi_{\alpha}^k=0$. So the same indices for which $\psi_{\alpha}^{k-1}=0$ on $W$, satisfy $\psi_{\alpha}^k=0$ on $O$. The remaining indices $\{\alpha_i:\ 1\le i\le n\}$ are then the only ones that can intersect $O$. This proves that the supports of the $\psi_{\alpha}^k$ form a locally finite family.