Well, this is embarassing. To be honest, during my PhD, I haven't really bothered too much regarding the topology of CW complexes. Back then, I understood the first few pages of Hatcher's book (the construction is easy to follow, also the examples given there), and didn't really care much about it. When I revised my stuff, I read abit about the topological part of it (for instance Hatcher's Appendix/Bredon's book regarding the topological aspects), but didn't really understand it (it wasn't that important back then) so I didn't really bother.
Now its stabbing me in the back. So I decided to crack this open and get over with it.
Hence, I've been reading Lee's Introduction to Topological Manifolds book, Chapter 5. It was going well until I hit Proposition 5.4.
The confusing part is this :
(1) In the proof of (C) condition, he writes "Because $\overline{e}$ is compact, it is covered by finitely many such neighborhoods". I can not see how this implies the closure finiteness condition. That is, from this, how is it that $\overline{e}$ is contained in a union of finitely many cells?
(Hatcher/Bredon also makes a similar jump - Hatcher's Prop A.1 and Bredon's Prop 8.1)
So, basically what the author is trying to tell us is, that for any $x \in \overline{e}$, there is a neighborhood $U_x$ that intersects only a finitely many cells of $\mathcal{E}$. From the next sentence, "Because $\overline{e}$ is compact, it is....", it appears to me that the author is using the following fact : $U_x$ is contained in a union of finitely many cells.
Why should this be true? If its false, how do you prove the (C) condition? And if its True, how do you prove it?
(2) I can not prove the equality of the expression after "it follows that". (The pink line that I've marked)
To be precise, how is RHS $\subseteq$ LHS?
For any $A\subseteq X$, we have $$A=A\cap X=A\cap\bigcup_{e\in\mathcal{E}}e=\bigcup_{e\in\mathcal{E}}A\cap e=\bigcup_{e\in\mathcal{E}\colon\ A\cap e\neq\emptyset}A\cap e\subseteq\bigcup_{e\in\mathcal{E}\colon\ A\cap e\neq\emptyset}e.$$ In your question (1), we take $A=U_x$ and the union is finite by the choice of $U_x$, so this gives the fact you were needing. In question (2), the same reasoning but with the closures establishes $A=\bigcup_{e\in\mathcal{E}\colon\ A\cap\overline{e}\neq\emptyset}A\cap\overline{e}$. However, $W\setminus A=W\setminus(W\cap A)$ and then $$W\cap A=\bigcup_{e\in\mathcal{E}\colon\ A\cap\overline{e}\neq\emptyset}A\cap W\cap\overline{e}=W\cap\bigcup_{i=1}^nA\cap\overline{e_i}$$ by the choice of $W$, so $W\setminus A=W\setminus\left(\bigcup_{i=1}^nA\cap\overline{e_i}\right)$.