Rotman's Algebraic Topology Theorem 8.24 I'm trying to understand the proof of Theorem 8.24 of Rotman's Introduction to Algebraic Topology. Two steps in his proof don't seem very obvious to me:

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*(blue highlight) Why do we need the "X is Hausdorff" statement to show that $\bar{e} = \Phi^n_{\alpha}(D^n)$ for some $\alpha$?


*(green highlight) The inductive hypothesis says that $X^{n-1}$ has a weak topology determined by $\{\bar{e}: dim(e) \leq n-1\}$, but I am not sure how I can proceed onward by applying Lemma 8.16.
Thanks everyone in advance!
Here is the main theorem:

and here is his proof:


Lemma 8.15:

Lemma 8.16:

 A: To show a cellular space is a CW-complex, notice that the first paragraph of the proof illustrates the given cellular space $X$ is Hausdorff. Read $(c)$ for blue highlight and $(d)$ for green highlight.
Now, consider $\Phi_\alpha^n$ as defined in the statement and suppose as an inductive hypothesis that the subset $X^{n-1}$ of the cellular space $X$is a CW-complex. Indeed, $X^{0}$ is a CW-complex. Now, we show $X^{n}$ is a CW-complex. Recall the definition of CW-complex as given on page $198$.

$(a)$ Note that as a set $$X^{n}= X^{n-1}\sqcup\bigsqcup_{\alpha\in A_n}\Phi_\alpha^n\left(\Bbb D^n\backslash \Bbb S^{n-1}\right)\text{ by }(ii)\text{ and the lemma below}.$$ By induction hypothesis $X^{n-1}$ is a CW-complex, so $X^{n-1}$ can be written as disjoint union of cells.  Also, denoting $e^n_\alpha:=\Phi_\alpha^n\left(\Bbb D^n\backslash \Bbb S^{n-1}\right)$  we can say $\Phi_\alpha^n\big| \Bbb D^n\backslash\Bbb S^{n-1}\to e_\alpha^n$ is a homeomorphism using the Lemma $1$ and $(ii)$.

Lemma $1$: Let $A, B$ be two topological spaces and $A'$ be a closed
subset of the space $A$ and $g\colon A'\to B$ be a continuous map.
Then the quotient map $q\colon A\sqcup B\to A\coprod_g B:=\frac{A\sqcup
 B}{a'\sim g(a')}$ sends $A\backslash A'$ homeomorphically onto the
subset $q(A\backslash A')$ of $A\coprod_g B$ such that $q(A\backslash A')\cap q(B)=\varnothing$.

This shows the cellular space $X^n$ is a disjoint union of cells, actually $$X^{n}=\bigsqcup_{0\leq k\leq n}\bigsqcup_{\alpha\in A_k}\Phi_\alpha^k\left(\Bbb D^k\backslash \Bbb S^{k-1}\right).$$

$(b)$ Now $X^{n-1}$ is a CW-complex (induction hypothesis) says that
the map $\Phi_\alpha^k\colon \left(\Bbb D^k,\Bbb S^{k-1}\right)\to \left(e_\alpha^k\cup X^{k-1}, X^{k-1}\right)$ is a relative homeomorphism, i.e., $\Phi_\alpha^k\big| \Bbb D^k\backslash\Bbb S^{k-1}\to e_\alpha^k$ is a homeomorphism for  $k=0,...,n-1$ and $\alpha\in A_k$. Now, writing $X^n$ as above (expression at the very beginning) and using the Lemma $1$, we can say each $\Phi_\alpha^n\colon \left(\Bbb D^n,\Bbb S^{n-1}\right)\to \left(e_\alpha^n\cup X^{n-1}, X^{n-1}\right)$ is a relative homeomorphism, i.e., $\Phi_\alpha^n\big| \Bbb D^n\backslash\Bbb S^{n-1}\to e_\alpha^n$ is a homeomorphism where $\alpha\in A_n$.


Lemma $2$: A compact subset of a CW-complex is contained in a finite
sub-complex, in particular, a compact subset is contained in a union
of finitely many cells of the CW-complex.

For proof of Lemma $2$, see Lemma $8.18$ on the page $201$ of the same book.
$(c)$ Let $k\in \{0,...,n\}$ and $\alpha\in A_k$ and write $e^k_\alpha:=\Phi^k_\alpha(\Bbb D^k\backslash \Bbb S^{k-1})$. We need to check $X^n$ is closure finite, i.e., $\overline{e^k_\alpha}$ is contained in a finite union of cells of $X^{(n)}$. If $k\leq n-1$ then $e^k_\alpha$ is a cell of $X^{n-1}$, and by induction hypothesis that $X^{n-1}$ is a CW-complex, we are done. If $k=n$, note that $$\overline{e^n_\alpha}=e^n_\alpha\sqcup \Phi_\alpha^n(\Bbb S^{n-1}).$$ Since, $\Phi_\alpha^n(\Bbb S^{n-1})$ is a compact subset of $X^{n-1}$ which by induction hypothesis is a CW-complex, applying the Lemma $2$, $\Phi_\alpha^n(\Bbb S^{n-1})$ is contained in a finite union of cells of $X^{-1}$. Hence, $\overline{e^n_\alpha}$ is contained in a finite union of cells of $X^n$.

$(d)$ Here, we check that the topology of $X^n$ is the weak-topology induced by the closure of all cells of $X^n$.

Fact: Let $A, B$ be two topological spaces and $A'$ be a closed
subset of the space $A$ and $g\colon A'\to B$ be a continuous map.
Consider  the quotient map $\pi\colon A\sqcup B\to A\coprod_g
 B:=\frac{A\sqcup  B}{a'\sim g(a')}$. Then,
$C\subseteq_\text{closed}A\coprod_g B$ if and only if
$\pi^{-1}(C)\subseteq_\text{closed} A\sqcup B$ if and only if
$A\cap\pi^{-1}(C)\subseteq_\text{closed} A\text{ and
 }B\cap\pi^{-1}(C)\subseteq_\text{closed} B$.

Recall (see statement) $$X^{n}=\left(\bigsqcup_{\alpha\in A_n}\Bbb D^n_\alpha\right)\coprod_f\left(X^{n-1}\right),$$ where $\displaystyle f=\bigsqcup_{\alpha\in A_n}f_\alpha^{n-1}\colon\bigsqcup_{\alpha\in A_n}\Bbb S^{n-1}\to X^{n-1}$.
Using the above Fact, we can now say that $K\subseteq_\text{closed} X^{n}$ if and only if $K\subseteq_\text{closed} X^{n-1}$ and $\left(\Phi^{n}_\alpha\right)^{-1}(K)\subseteq_\text{closed} \Bbb D^n_\alpha$ for each $\alpha\in A_{n}$. Note that a little bit set-equality checking is needed (which is more or less obvious) with the observation that  $X^{n-1}\subseteq X^n$.
Now, $X^{n-1}$ is a CW-complex by inductive hypothesis, i.e., induction hypothesis says that $K\subseteq_\text{closed} X^{n-1}$ if and only if  $\left(\Phi^{m}_\alpha\right)^{-1}(K)\subseteq_\text{closed} \Bbb D^m_\alpha$ for $0\leq m\leq n-1$ and $\alpha\in A_m$.
Hence, $$ K\subseteq_\text{closed} X^n\iff \left(\Phi^{m}_\alpha\right)^{-1}(K)\subseteq_\text{closed} \Bbb D^m_\alpha\text{ for }0\leq m\leq n.$$ Therefore, $$\bigsqcup_{0\leq k\leq n}\bigsqcup_{\alpha\in A_k}\Phi_\alpha^k\colon\ \bigsqcup_{0\leq k\leq n}\bigsqcup_{\alpha\in A_k}\Bbb D_\alpha^k\to X^n$$ is a quotient map.
Now, applying Lemma $8.16$, we are done.
