So I'm, trying to prove that a cell complex is Hausdorff. I've looked at a few proofs (namely that of Hatcher's algebraic topology) but a lot of them prove instead that a cell complex is normal. I just want to do the most basic and working from the definitions so I can understand it, but there are a lot of things I'm not comfortable with just yet that I wanted to clear up. So I defined a cell complex as a union of $X^n$ where
$$X^n = \{X^{n-1} \sqcup (\sqcup_\alpha D_{\alpha}^{n})\}/ \sim$$
where $\sim$ represents the relation $s \sim f_{\alpha}^n(s)$ where $ s \in \partial D_{\alpha}^n$ and $f_{\alpha}^n: \partial D_{\alpha}^n \to X^n$ is the attaching map of the $\alpha$-th n- cell.
Now I know that there is this quotient map $q: \{X^{n-1} \sqcup (\sqcup_\alpha D_{\alpha}^{n})\} \to \{X^{n-1} \sqcup (\sqcup_\alpha D_{\alpha}^{n})\}/ \sim $ and from what I understand about quotient spaces a set $U \subset \{X^{n-1} \sqcup (\sqcup_\alpha D_{\alpha}^{n})\}/ \sim $ is open if and only if $U$ is open in $\{X^{n-1} \sqcup (\sqcup_\alpha D_{\alpha}^{n})\} $. Now I'm sort of having trouble seeing what an open set in $\{X^{n-1} \sqcup (\sqcup_\alpha D_{\alpha}^{n})\} $ looks like. I understand that these cell complexes are build recursively. So I was thinking that open in $\{X^{n-1} \sqcup (\sqcup_\alpha D_{\alpha}^{n})\} $ means open in $\{X^{n-2} \sqcup (\sqcup_\alpha D_{\alpha}^{n-1})\} $. But then if you follow this recursively you eventually get back to $X^0$ which is a set of finite points. Which I don't really understand how to define as open, is it the subspace topology on $X$? I'm not sure.
Then I also saw that $X$ has the weak topology where $A \subset X$ is open in $X $ if and only if $A \cap X^n$ open in $X^n$ for all $n$, but then that leads me back to what does an open set in $X^n$ look like.
I also saw that for each attaching map $f_{\alpha}^n$ you have this extension called the characteristic map $\Phi_{\alpha}^n: D_{\alpha}^n \to X$ and I saw that $A \subset X$ is open if and only if $(\Phi_{\alpha}^n)^{-1}(A)$ is open in $D_{\alpha}^n$ for all $\alpha$ and for all $n$. But then what does it mean to be open in $D_{\alpha}^n$, is this the subspace topology on X? And eventually $D_{\alpha}^0$ are a bunch of points (0-cells) so how can something be open in that, even using the subspace topology I cannot see it.
After I understand that I want to show the Hausdorff property by taking $a,b \in X$ and finding some neighborhoods $N_a, N_b$ respectively so that $N_a \cap N_b = \emptyset$. But I am just having trouble understanding what these neighborhoods would look like.
I know this is super long, but I would really like it if somebody could explain it to me, I've been trying to use various texts and online references but I still do not fully understand.