Let $X$ be a $n$-dimensional CW complex with $n \geq 1$, and $e^n_{\alpha}$ a n-cell of $X$. The clain is show that $X - e^n_{\alpha}$ is a subcomplex of $X$ and $X$ is homeomorphic to an adjunction space obtained from $X - e^n_{\alpha}$ by attaching a single $n$-cell.
So, to show that $X - e^n_{\alpha}$ is a subcomplex, we have to show that $X - e^n_{\alpha}$ is the union of cells of $X$, such that if $e \in X - e^n_{\alpha}$ then $\bar{e} \in X - e^n_{\alpha}$. The first is easy, because we are just taking a single cell of $X$. So $X - e^n_{\alpha}$ is a union of cells of $X$. The second clain follows because if $e \in X - e^n_{\alpha}$ then $e\in X$ and $e\cap e^n_{\alpha} = \emptyset $ i.e $e^n_{\alpha} \neq e$, so $\bar{e} \cap e^n_{\alpha} \subset \bar{e}-e$ which is contained in a union of finitely many cells of dimensions less than $n$, since $e^n_{\alpha}$ has dimension $n$, we have $e^n_{\alpha}\cap \bar{e}$ is empty, so $\bar{e} \in X-e^n_{\alpha}$.
The second clain is easy to see geometrically, but I have truble to organize my thoughts. Assuming that $X = X_{n-1} \cup_{\phi} (\sqcup_{\alpha} D_{\alpha}^n)$, remove a single cell is equivalent to remove a single disc in $X_{n-1} \sqcup (\sqcup_{\alpha} D_{\alpha}^n)$. I will denote this space as $X_{n-1} \sqcup (\sqcup_{\alpha'} D_{\alpha'}^n)$. If I proof that $q : X_{n-1} \sqcup (\sqcup_{\alpha'} D_{\alpha'}^n \cup D^n) \to X$ is a quotient map that's make the same identification as a quotient map in $X_{n-1} \cup_{\phi} (\sqcup_{\alpha} D_{\alpha}^n)$ I've done, right?
My ideas are correct? I will appreciate any suggestion.