In the third paragraph on pg. 7 of Hatcher's Algebraic Topology there is the definition of characteristic maps of a CW complex (the definition of a CW complex can be found on pg. 5.)
Just after defining a characteristic maps, Hatcher writes:
The restriction of a characteristic map $\Phi_{\alpha}:D^n_\alpha\to X$ to the interior of $D^n_\alpha$ is a homeomorphism onto $e^n_\alpha$ (where $e^n_\alpha$ is the image of $\text{Int}(D^n_\alpha)$ under $\Phi_\alpha$).
In, other words, the restriction of $\Phi_\alpha$ to the interior of $D^n_\alpha$ is an embedding.
I was able to prove this by assuming additionally that $X$ is Hausdorff. Here is the argument I have:
Lemma. Let $B^n$ be the open unit ball in $\mathbf R^n$. Let $X$ be a Hausdorff topological space and $f:D^n\to X$ be a continuous map such that
i) $f$ is injective on $B^n$, and
ii) $f(S^{n-1})\cap f(B^n)$ is empty.
Then $f|_{B^n}:B^n\to X$ is a topological embedding.
Proof. Without loss of generality we may assume that $f$ is surjective. We have the following diagram:
where by $D^n/f$ we mean the quotient space of $D^n$ formed by the fibres of $f$. We need to show that $f|_{B^n}:B^n\to X$ is an embedding. Since $f$ is a surjective continuous map from a compact space to a Hausdorff space, we know that $f$ is a quotient map. Therefore $\bar f$ is a homeomorphism. In the light of this, it suffices to prove that $\pi|_{B^n}:B^n\to \pi(B^n)$ is a homeomorphism. To this end, note that since $f$ is injective on $B^n$, we have $\pi|_{B^n}:B^n\to \pi(B^n)$ is bijective. So all that remains to show is that $\pi|_{B^n}$ is an open map.
To this end, let $U$ be open in $B^n$. We claim that $\pi^{-1}(\pi(U))=U$. One containment is obvious. For the reverse containment let $x\in \pi^{-1}(\pi(U))$. Then $\pi(x)\in \pi(U)$, and thus $\bar f\circ \pi\in \bar f\circ \pi(B^n)$, giving $f(x)\in f(B^n)$. Using (ii) here leads to the conclusion that $x\in B^n$. But since $f$ is injective on $B^n$, we further deduce that $x\in U$. So we have $U=\pi^{-1}(\pi(U))$.
From this we infer that $\pi(U)$ is open in $D^n/f$ and is therefore open in $\pi(B^n)$ too.
Remark. The condition (ii) is necessary in the hypothesis of the above proposition for result to hold. For otherwise we have the figure-eight curve is would serve as a counterexample.
It is clear that a characteristic map satisfies the hypothesis of the above lemma and thus given an embedding when restricted to the interior of the disc.
Question. How do we prove that the restriction of a characteristic map to the interior of the disc is an embedding without assuming that $X$ is a Hausdorff space.
We have the following general fact
Assume $X$ is space, and $f:A\to Y$ is a map from the subspace $A$ of $X$ to a space $Y$. We have the following pushout square.
Then if $A$ is closed (open) in $X$, the restriction $g:X\setminus A \to Z$ of $F$ is a closed (open) embedding.
You already gave a proof for the case that $A$ is closed (by replacing the pair ($D^n,S^{n-1})$ in your proof with a general $(X,A)$), and for the case of an open $A$ simply switch the terms open and closed in the proof.
Now the $n$-skeleton $X^n$ is constructed as a pushout
where $\varphi$ is the union of all attaching maps $\varphi_\alpha$. Since $\coprod S_\alpha^{n-1}$ is closed in $\coprod D_\alpha^n$, it follows that $\Phi$ restricts on each open ball $B_\alpha^n$ to an embedding.