I'm studying CW-complexes and regular CW-complexes be a special case . It has a basic property as you can see in lemma $5.5$ in the picture above . I assume you are familiar with CW-complexes . Could you explain the yellow line in the picture above . Invariance of domain just gives me $e^{n-1} \cap \overline{e}^{n}$ is open in $\overline{e}^{n} - e^{n}$ , not $e^{n-1}$
Let $\varphi:S^{n-1}\to K^{n-1}$ be the the attaching map for the cell $e^n$. Recall that $e^{n-1}$ is an open subset of $K^{n-1}$, thus the preimage $P=\varphi^{-1}(e^{n-1})$ is an open subset of $S^{n-1}$. Notice also that $P\neq S^{n-1}$ for compactness reasons, and so, using stereographic projection, we may think of $P$ as an open subset of $\Bbb R^{n-1}$.
Since $\varphi$ is injective, and thus induces a homeomorphism onto its image for compactness / Hausdorff reasons, it restricts to a homeomorphism from $P$ to $\varphi(S^{n-1})\cap e^{n-1}$, or $\dot{e}^n\cap e^{n-1}$ in Massey's notation. However, recall that the subspace topology $e^{n-1}$ inherits from $K^{n-1}$ coïncides with the standard topology on $e^{n-1}=D^{n-1}$
Hence, $\varphi$ induces a homeomorphism from $P$ to $\varphi(S^{n-1})\cap e^{n-1}$, a certain subset of $e^{n-1}\simeq D^{n-1}$. In particular, it induces a continuous injection from $P$ to $D^{n-1}$, both open subsets of $\Bbb R^{n-1}$.
By invariance of domain, an injective continuous map between open subsets of $\Bbb R^n$ is open, and hence $\varphi(S^{n-1})\cap e^{n-1}$ is open.