Below is the proof of the Open Mapping Theorem from Rudin's Functional Analysis. Near the end of the proof, I cannot figure out why $y_{m+1} \to 0$ as $m \to \infty$ by the continuity of $\Lambda$. I think this has something to do with the fact that $y \in \cap_n \overline{\Lambda{V_n}}$, but I cannot show this. I would greatly appreciate any help.
I think I have solved this question. The idea is to use that $y_n \in \overline{\Lambda(V_n)}$ for all $n$, where $V_n = B_d(0,2^{-n}r)$.
First, note that if we have any sequence $x_n \in X$ such that $x_n \in V_n$ for all $n$, then $x_n \to 0$. This is because, for all $\epsilon>0$, we can choose an $N$ such that $2^{-N}r < \epsilon$, so that for $n \ge N$, $d(x_n,0) < 2^{-n}r < 2^{-N}r < \epsilon$.
Therefore, for any sequence $\Lambda(x_n) \in \Lambda(V_n)$, by continuity of $\Lambda$, we have $\Lambda(x_n) \to 0$. Now we use that for any $y_n$, there is some $y_n' \in \Lambda(V_n)$ close to $y_n$, so that $y_n$ must converge to $0$ as well.
Take any neighborhood of $0$, say $U$. Then take a symmetric neighborhood $V$ of $0$ such that $V+V \subset U$. This can be done as in Theorem 1.10 of Rudin 's Functional Analysis.
Then since $y_n$ is in the closure of $\Lambda(V_n)$, for each $n$, we have $y_n + V \cap \Lambda(V_n) \neq \emptyset$. So take $y_n ' = \Lambda(x_n')$ from the intersection. Then from the previous argument, we have $y_n' \to 0$, so that there is some $N$ such that if $n \ge N$ then $y_n' \in V$. Meanwhile, we also have $y_n - y_n' \in V$ for all $n$. Therefore, if $n\ge N$ we have $y_n = y_n-y_n'+y_n' \in V + V \subset U$.
Since $U$ is an arbitrary neighborhood of $0$, we have $y_n \to 0$.