From the proof of Proposition 5.18 given below, I can't understand the last paragraph. Why does the assumption that $A \cap X_{n-1}$ is closed in $X_{n-1}$ imply that $B \cap \bar{e}$ is closed in $\bar{e}$ for every cell $e$ of dimension less than $n$; and $A \cap D_\alpha^n$ is closed in $D_\alpha^n$ for each $\alpha$ imply that $B \cap \bar{e_\alpha^n}$ is closed in $\bar{e_\alpha^n}$?
The definition of a CW complex is given as follows.
To prove that $\Phi$ is a quotient map, we have to show that if $B \subset X_n$ is a set such that $\Phi^{-1}(B)$ is closed in $S = X_{n-1} \sqcup \bigsqcup_\alpha D^n_\alpha$, then $B$ is closed in $X_n$. Note that $\Phi_\alpha = \Phi \mid_{D^n_\alpha} : D^n_\alpha \to X_n$ is the characteristic map for the cell $e^n_\alpha$.
$\Phi^{-1}(B)$ closed means that $\Phi^{-1}(B) \cap X_{n-1}$ is closed in $X_{n-1}$ and $\Phi^{-1}(B) \cap D^n_\alpha$ closed in $D^n_\alpha$, hence compact, for all $\alpha$.
For any $A \subset S$ we have $\Phi^{-1}(B) \cap A = \Phi^{-1}(B \cap \Phi(A)) \cap A = (\Phi \mid_A)^{-1}(B \cap \Phi(A))$.
For $A = X_{n-1}$ we have $\Phi(A) = X_{n-1}$ and $\Phi\mid_{X_{n-1}}$ is the inclusion of $X_{n-1}$ into $X_n$. Hence $\Phi^{-1}(B) \cap X_{n-1} = (\Phi\mid_{X_{n-1}})^{-1}(B \cap X_{n-1}) = B \cap X_{n-1}$, thus $B \cap X_{n-1}$ is closed in $X_{n-1}$ which implies that $B \cap \overline e$ is closed in $\overline e$ for all cells of dimension $< n$.
For $A = D^n_\alpha$ we have $\Phi(A) = \overline{e^n_\alpha}$. Hence $\Phi^{-1}(B) \cap D^n_\alpha = \Phi_\alpha^{-1}(B \cap \overline{e^n_\alpha})$ is compact, thus also $\Phi_\alpha(\Phi_\alpha^{-1}(B \cap \overline{e^n_\alpha}))= B \cap \overline{e^n_\alpha}$ is compact and therefore closed in $\overline{e^n_\alpha}$.