I think that you are correct in the hint that you're remembering that the exercise told you to consider.
My idea is that, using this hint, you ought to be able to proceed inductively in constructing the regular complex $Y$.
Start by looking at all attaching maps $\varphi_{\alpha}:S^{\alpha} \rightarrow X_0$ onto the $0$-skeleton for $X$. Since $X$ is finite, these can be listed: $\{\varphi_{\alpha_1},...,\varphi_{\alpha_r}\}$.
Now, as you suggest, for each of these we wish to somehow consider instead the map $S^{\alpha} \rightarrow X_0 \times D^{\alpha+1}$, because this would be injective. What we therefore do is, for each $\alpha_1,...,\alpha_r$, take the successive product of $X_0$ with each of the $D^{{\alpha_i}+1}$s. This results in a space
$$Y_0 := X_0 \times D^{{\alpha_1}+1} \times....\times D^{{\alpha_r}+1}.$$
We can now use the standard cellular decomposition of $D^{k+1}$ to obtain a cellular decomposition of the product above. This is the "bottom" of our new regular complex $Y$. It is not the $0$-skeleton for our CW-complex $Y$, but it is homotopy equivalent to the $0$-skeleton of $X$. As such, we do indeed mildly abuse notation by calling this thing $Y_0$.
We then proceed with attaching cells to this product via the maps $\psi_{\alpha}$ as you suggest above. Importantly, we proceed as we normally would when we construct a CW-complex. We first attach the $1$-cells, and procede with $2$-cells, etc.
It is important to note a couple of things at this point.
We have certainly dealt with all attaching maps
for the $1$-skeleton of $X$ at this point. What this means is that
this process can indeed now be iterated; we look at attaching maps onto $X_1$, take whatever products we need to take with disks D^{\alpha}$ in order that these maps will be injective, and continue.
Importantly what we have constructed (and what we construct inductively in
subsequent steps) is homotopy equivalent to $X_0$, $X_1$ and so on.
It is indeed easy enough to see that the product $X_0 \times
D^{{\alpha_1}+1} \times....\times D^{{\alpha_r}+1}$ is homotopy
equivalent to $X_0$, but what about the next steps? We wish to see that $Y_1$ is homotopy equivalent to $X_1$. This can be seen in a few different ways, but the simplest for these purposes is just to state the following lemma (which I state rather wordily):
Lemma: Due to the nature of the construction of CW-complexes,
whatever space $X$ you're attaching a particular cell to can be replaced by a
homotopy equivalent space $\overline{X}$ (and the attaching map by the appropriate homotopic map) and the space which results $\overline{X} \cup D^{k}$ is homotopy equivalent to $X \cup D^k$.
Now what we're essentially doing is replacing $X_0$ with the homotopy equivalent $Y_0$, and attaching each $1$-cell to this new thing. Utilising the above lemma for each cell we attach gives that the resulting space is homotopy equivalent to $X_1$. We now "fatten up" this resulting space in exactly the same manner as how we "fattened up" $X_0$, by taking products with disks. Taking products with disks trivially results in a homotopy equivalent space, and what we end up with is the space which we will label $Y_1$; homotopy equivalent to $X_1$.
I think that iteratively doing this should complete the proof.
I should note that whilst the $Y_0$ and $Y_1$ which I mention above are certainly not the $0$ or $1$ skeletons for the space $Y$, there is nevertheless a natural $CW$-complex structure available for $Y$ which is obtained by using the standard cellular decomposition of the disks $D^k$, and this is then indeed a regular complex.
Indeed, the component of $q^{-1}(A\cap X^n)$ in $X^{n-1}$ is $A\cap X^{n-1}$. This is essentially by definition, but let us spell it out, because Hatcher is not always explicit about this. In reality, $X^{n-1}$ is not a literal subset of $X^n$, instead we always canonically identify with a subset of $X^n$ by the composition $X^{n-1}\hookrightarrow X^{n-1}\sqcup\coprod_iD_i^n\twoheadrightarrow X^n$, where the first map is the inclusion and the second map is the quotient map and $A\cap X^{n-1}$ is precisely the preimage of $A$ (when considered a subset of $X^n$) under this map, i.e. it is the component of $q^{-1}(A\cap X^n)$ in $X^{n-1}$.
Now, fix an index $i$ and let $\Phi_i\colon D_i^n\rightarrow X$ be the characteristic map of $D_i^n$ and $e_i^n\subseteq X$ the corresponding $n$-cell. There are two options: 1.) $e_i^n\subseteq A$. Then $\overline{e_i^n}\subseteq A$ by definition of subcomplex, whence $\phi_i(D_i^n)=\phi_i(\overline{\mathrm{int}(D_i^n)})\subseteq\overline{\phi_i(D_i^n)}=\overline{e_i^n}\subseteq A$, so that the component of $q^{-1}(A\cap X^n)$ in $D_i^n$ is precisely $\phi_i^{-1}(A)=D_i^n$, so in particular closed. 2.) $e_i^n\cap A=\emptyset$ (this is a dichotomy, because we know that $A$ is a union of cells and the cells are pairwise disjoint). In this case, the component of $q^{-1}(A\cap X^n)$ in $D_i^n$ consists only of points on the boundary $\partial D_i^n$ that get mapped to $A$, but those are precisely the points in the preimage $\varphi_i^{-1}(A\cap X^{n-1})$, where $\varphi_i$ is the attaching map of $D_i^n$, so in particular closed. All in all, $q^{-1}(A\cap X^n)$ is closed in $X^{n-1}\sqcup\coprod_iD_i^n$ and the argument goes through.
Best Answer
It is extremely, extremely easy to show a CW-complex is Hausdorff, include it in your proof if you are worried about it.
With this fact, the closure of an open cell $e \rightarrow X$ is the image of $e \cup S^n \rightarrow X$ given by the inclusion of the open cell and the characteristic map on the boundary. This is because $e \cup S^n = D^{n+1}$ is compact, and the image of a compact set is compact which in a Hausdorff space implies closed. This is the smallest closed set containing the image of $e$ since any point in the image of the characteristic map is in the boundary of the image of $e$.