Thomas has already explained the answer to your question. For my answer, it does not answer your question directly but it provides another reason why each $n$ -cell $e_\alpha^n$ is open in $X$.
You already know that your $n$ -cells are attached via a characteristic map $\varphi_\alpha : S^{n-1} \to Y$. $Y$ now is the space that when you attach $n$ -cells $e_\alpha^n$ you get $X$. Now each $e_\alpha^n$ has its characteristic map $\Phi_\alpha$ which is defined to be the composition
$$D_\alpha^n \hookrightarrow Y \sqcup D_\alpha^n \to Y \sqcup D_\alpha^n/\sim $$
where the quotient space is just $X$. Recall that $\sim$ is the smallest equivalence relation generated by the subset of all pairs $(x,\varphi(x))$ for $x \in S^{n-1}$ in
$$(Y \sqcup D_\alpha^n) \times (Y \sqcup D_\alpha^n).$$
Now it is not hard to see that a subset $A$ of $X$ is open iff $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for each characteristic map $\Phi_\alpha$. Now suppose you have an $n$ -cell $e_\alpha^n$. Then if $\alpha \neq \beta$, then it is clear that $\Phi^{-1}_\beta(e_\alpha^n ) = \emptyset$ that is open in $D_\beta^n$. Now if $\beta = \alpha$, this is trivial to see because we would have that $\Phi$ is a homeomorphism of the interior of $D_\alpha^n$ and $e_\alpha^n$.
Is it clear to you why an $n$ - cell is open now?