Let $A\subset X^0$. For any cell $c\in\mathcal{C}$, let $X_c$ be a finite subcomplex that contains $\bar{c}$. Note that $A\cap X_c=X_c\setminus(X_c\cap A)$. As $X_c$ is finite, we have $X_c\setminus(X_c\cap A)$ is finite. Additionally, $X$ is Hausdorff, $A\cap X_c=X_c\setminus(X_c\cap A)$ is closed because finite set is closed in Hausdorff space. Thus, $A$ is closed by the weak topology.