This is a statement from Hatcher's Algebraic Topology that I can't completely understand. So we define a CW complex $X$ inductively starting with a discrete set $X^0$ the $0$-cells of $X$. Then inductively, form the $n$-skeleton $X^n$ from $X^{n-1}$ by attaching the $n$-cells $e_\alpha^n$ via maps $\phi_\alpha:S^{n-1}\to X^{n-1}$. This means that $X^n$ is the quotient space of $X^{n-1} \coprod_\alpha D_\alpha^n$ under the identifications $x \sim \phi_\alpha(x)$ for $x \in \partial D_\alpha^n$. The cell $e_\alpha^n$ is the homeomorphic image of $D_\alpha^n- \partial D_\alpha^n$ under the quotient map.
Finally, $X=\cup_n X^n$ with the weak topology: A set $A \subset X$ is open(closed) iff $A\cap X^n$ is open(closed) in $X^n$ for each $n$.
Then the text says an alternative way to describe the topology on $X$ is to say that a set $A \subset X$ is open(or closed) iff $\Phi_\alpha^{-1}(A)$ is open(or closed) in $D_\alpha^n$ for each characteristic map $\Phi_\alpha$.
One direction follows from continuity of the $\Phi_\alpha's$ and in the other direction, suppose $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for each $\Phi_\alpha$, and suppose by induction on $n$ that $A \cap X^{n-1}$ is open in $X^{n-1}$. Then since $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for all $\alpha,$ $A \cap X^n$ is open in $X^n$ by the definition of the quotient topology on $X^n$.
I can't follow the last sentence. Why does the definition of quotient topology on $X^n$ imply that $A \cap X^n$ is open in $X^n$ if $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for all $\alpha$?
Thinking about this for a while I think it is true because of the following argument. Inductive construction gives the map $$\Phi: X^{n-1} \coprod (\coprod_\alpha D_\alpha^n) \to X^n$$ that is equal to inclusion on $X^{n-1}$ and to $\Phi_\alpha$ on each $D_\alpha^n$ makes the same identifications as the quotient map $\Phi_\alpha$.
We have $A \cap X^n=(A\cap X^{n-1})\coprod_\alpha(A \cap(D_\alpha^n)^o). $ By the definition of quotient topology, $A\cap X^n$ is open in $X^n$ if $\Phi^{-1}(A\cap X^n)$ is open in the disjoint union topology. And by the topology of the disjoint union topology, $\Phi^{-1}(A\cap X^n)$ is open if $\Phi^{-1}(A\cap X^n)\cap X^{n-1}$ is open in $X^{n-1}$ and $\Phi^{-1}(A\cap X^n)\cap \coprod D_\alpha^n$ is open in $\coprod_\alpha D_\alpha^n$. But $\Phi^{-1}(A\cap X^n)\cap X^{n-1} = A\cap X^{n-1}$ which is open in $X^{n-1}$ by induction hypothesis. And $\Phi^{-1}(A\cap X^n)\cap \coprod D_\alpha^n=\coprod_\alpha \Phi_\alpha^{-1}(A)\cap D_\alpha^n$ which is open in $\coprod_\alpha D_\alpha^n$ by assumption. Hence, $\Phi^{-1}(A\cap X^n)$ is open in the disjoint union topology so $A\cap X^n$ is open in $X^n$ by the quotient property.