A question about the proof of the open mapping theorem (2.11) in Walter Rudin's Functional Analysis Can refer to page 65-66 of this pdf.
I actually have two questions about the proof. The first is, in proving the first part of (2), why $\overline{\Lambda (V_2)}$ contains a neighborhood of $0$? I think it only proves $\overline{\Lambda (V_2)}$ has a nonempty interior. It is not quite obvious to me why this interior must contain $0$.
My second question is, in proving the second part of (2), why $y_{m+1} \to 0$? I don't quite see how this can be inferred from the continuity of $\Lambda$.
 A: For your first question: For any $k\in\mathbb{N}$, $V_r=\{x\in X: d(x,0)<2^{-k}r\}$, where $r>0$ is  fixed.
$\overline{\Lambda(V_2)}$ has nonempty interior. Denote it by $W$. Let $x,y\in V_2$. Then
$$d(x-y),0)=d(x,y)\leq d(x,0)+d(0,y)<2^{-2}r +2^{-2}r=2^{-1}r$$
That is $V_2-V_2\subset V_1$; hence
$$0\in W-W\subset \overline{\Lambda(V_2)}-\overline{\Lambda(V_2)}\subset\overline{\Lambda(V_2)-\Lambda(V_2)}\subset\overline{\Lambda(V_1)}$$
$W-W$ is open neighborhood of $0$, and contained in $\overline{\Lambda(V_1)}$.
(The same argument works for $V_k$ ($k>2$) in place of $k=2$).

For your second question: notice that $y_n\in\overline{\Lambda(V_n)}$ for all $n$. Let $U$ be any neighborhood of $0$ in $Y$ and $G$ be symmetric open neighborhood of $0$ such that $G+G\subset U$. Choose $z_n\in V_n$ such that
$y_n-\Lambda z_n\in G$. Clearly $z_n\xrightarrow{n\rightarrow\infty}0$ in $X$ since  $d(z_n,0)<2^{-n}r\xrightarrow{n\rightarrow\infty}0$. The continuity of $\Lambda$ implies that for some $N$, $n\geq N$ implies that $\Lambda z_n\in G$. Putting things together, for $n\geq N$,
$$y_n=\big(y_n-\Lambda z_n\big) +\Lambda z_n\in G+G\subset U$$
This means that $y_n\xrightarrow{n\rightarrow\infty}0$ in $Y$.
