I am reading the proof of 6.5 Theorem in Rudin's book. For part (c), he wants to prove that a bounded subset $E \subset \Omega$ must be contained in a $\mathscr{D}_{K}$ for some compact subset $K \subset \Omega$. A bit awkwardly that I don't understand one of his argument, at the bottom paragraph of page 154, which says
Since each K contains only finitely many $x_{m}$, it is easy to see that $\mathscr{D}_{K} \cap W \in \tau_{K}$. Thus $W \in \beta$.
Sadly it seems not easy to me. I cannot figure out why $\mathscr{D}_{K} \cap W \in \tau_{K}$ for each $K$. I know what $\phi_{m}$ and $x_{m}$ are, and know why $K$ contains finitely many $x_{m}$ (,since if not, there must be an accumulation point. Am I right?). The trouble is I don't understand the "easy to see" stuff:-(
Thanks
$\mathscr D_K \cap W = \bigcap_{x_{j}\in K} \lbrace\varphi \in \mathscr D_K: |\varphi(x_j)| <c_j\rbrace$ for some finitely many $x_{m}$ in $K$ and corresponding constants $c_j>0$ (given explicitly in Rudin's proof). So all you need is the continuity of the evaluation $\mathscr D_K\to \mathbb C$, $\varphi\mapsto \varphi(x)$.
By the way, I do not like Rudin's approach here. I find it conceptually more instructive to topologize $\mathscr D(\Omega)$ as the inductive limit of $\mathscr D_K$ (which are closed subspaces of $C^\infty(\Omega)$). A proof that in such a situation every bounded set of the inductive limit is contained in some "step" is not longer than in the concrete situation.