Homotopy equivalence involving cell - complexes I have two small questions related to some point-set topology involved in the following. My  question is inspired after reading the proof of Proposition 1.26 of Hatcher.
Suppose I have a space $X$ that is built out of a subspace $A$ that is path-connected by attaching some $n$ - cells  $e_\alpha^n$ for $n \geq 3$ via attaching maps $\varphi_\alpha : S^{n-1} \to A$. Suppose for each $\alpha$ I choose a point $y_\alpha \in D_\alpha$. Now suppose I set 
$$U = X - \left\{\bigcup_{\alpha} y_\alpha\right\}.$$
I am interested to know why $(1)$ $U$ is open in $X$. It could be that a union of infinitely many point sets is not closed, from which we cannot conclude that $U$ is open. How can I see this fact?
My second question is somewhat related to the proposition as well in that I believe:


$U$ is homotopy equivalent to my subspace $A$ from which $X$ is built out of.


Now I don't know if this is true and I am trying to prove it. I already have an inclusion map $i : A \to U$ to use. However, I don't necessarily have a map going the other way from $U$ to $A$. Is my belief true and if it is, how would I go along proving it? At the end of the day, I would like to be able to fill in the details of this by myself.
Thanks.
Edit: I should say that my $\alpha$'s run over an arbitrary index set, not necessarily countable or anything.
 A: I'm first going to answer the question I think you meant to ask. If I'm mistaken, please say so and I'll remove this.
Looking at the section in Hatcher you indicated, I think you meant your $D_\alpha$ to be open cells, i.e. images of open disks with the characteristic maps (this is what Hatcher denotes $e^n_\alpha$). In this case it is simple to see that $U$ is open. Recall that a subset of a CW complex is open/closed iff it intersects each closed cell in an open/closed set (this is a defining characteristic of cell complexes; Hatcher discusses this in the appendix, I think). The intersection of your set $U$ and a closed cell in $X$ is simply the cell, minus an interior point. Pulling back to the disk, this is clearly open, so $U$ must be open.
Similarly, if your $D_\alpha$ are open cells, the answer to your second question is yes. To prove this, it suffices to show that $U$ deformation retracts onto $A$. Let me give a quick sketch of this in the case there is only one (aka finitely many) cell(s). We have removed a point from the interior of the single cell. Pulling this cell back, its preimage with its characteristic map is a punctured disk (remember that the characteristic map is a homeomorphism when restricted to the interior of the disk). There is an obvious deformation retraction of this onto the boundary $S^1$. Now just take this deformation retraction and combine it with the characteristic map to get a deformation retraction of the (closed) cell onto its boundary in $X$. Since this boundary lies in $A$, we are done.
As I said, the same works for finitely many cells. When dealing with infinitely many cells, you might need to use a (fairly standard) contrivance to be able to contract all of the cells in "finite time". This method is also shown in Hatcher, probably somewhere around the point where he introduces $S^\infty$. 

If you insist on having $D_\alpha$ be closed cells, I expect things can go wrong. I think $U$ might not be open in this case, but I'm too rusty at the moment to think of a counterexample.
Certainly what you conjecture in your second question is false in this case. Take a closed interval as $A$ and attach an $n$-cell at one of the boundary points. Let that boundary point be the point $y$ you remove. Then $U$ isn't even connected and can't be homotopy equivalent to $A$.
