Let $X = \bigcup\limits_{n=0}^{\infty} X^n$ be a CW-complex. Let $A \subseteq X$ be such that $A$ pulls back to an open set under the characteristic map $\varphi_i : D^n_i \longrightarrow X.$ Show that $A$ is open in $X$ with respect the weak topology.
Proof $:$ It is clear that $A \cap X^0$ is open since $X^0$ is a discrete set. Suppose that $A \cap X^{n-1}$ is open in $X^{n-1}.$ Now $X^n$ is a quotient space of $X^{n-1} \bigsqcup \left ( \coprod\limits_i D^n_i \right ).$ Since $A \cap X^n$ pulls back to open sets in each piece of this coproduct under the quotient map it turns out that $A \cap X^n$ is open in $X^n$ by the definition of the quotient topology and hence by induction hypothesis we are done with the proof.
In the above proof I don't understand why $A \cap X^n$ pulls back to an open set under the quotient map in each piece of the coproduct. Is the pull back of $A \cap X^{n}$ lying in $X^{n-1}$ is $A \cap X^{n-1}\ $? It is clear that the pull back $A \cap X^n$ lying in $D_i^n$ under the quotient map is same as the inverse image of $A$ under the characteristic map $\varphi_i.$ Since $\varphi_i^{-1} (A)$ is given to be open in $D_i^n$ it follows that the pull back of $A \cap X^n$ lying in $D_i^n$ under the quotient map is open in $D_i^n.$ So if we can show that $A \cap X^n$ pulls back to $A \cap X^{n-1}$ in $X^{n-1}$ under the quotient map then we are through as we already know that $A \cap X^{n-1}$ is open in $X^{n-1}.$ We need to find out the inverse image of $A$ under the map obtained by composing the following maps $$X^{n-1} \hookrightarrow X^{n-1} \bigsqcup \left (\coprod\limits_i D_i^n \right ) \xrightarrow {\text {quotient map}} X^n \hookrightarrow X.$$
Would anybody please help me in this regard? Thanks for your time.
EDIT $:$ Since $X^{n-1} \subseteq X^{n}$ I think that the composition of the first two maps actually gives an inclusion because as a set $X^{n} = X^{n-1} \bigsqcup \left (\coprod\limits_i e^n_i \right ),$ where $e^n_i$ is the $i$-th open $n$-cell. So $A \cap X^{n}$ pulls back to $A \cap X^{n} \cap X^{n-1} = A \cap X^{n-1}$ in $X^{n-1},$ which is open $X^{n-1}$ by induction hypothesis. Is it what is happening here?