In the proof of Theorem 5.4 in Rudin RCA (Real and Complex Analysis), the following conclusion is reached.
Specifically regarding the portion that (c) implies (a). That is if $\Lambda$ is continuous at one point of $X$ then $\Lambda$ is bounded.
I am not sure how the above was arrived at. Attached below is the full theorem and proof of Theorem 5.4 - with all the relevant symbols and terminology - from Rudin RCA.
My attempt is given below to complete the last step in the proof from Rudin.
$$ \left\Vert \Lambda x\right\Vert <\epsilon $$ $$ \Rightarrow\frac{\left\Vert \Lambda x\right\Vert }{\delta}<\frac{\epsilon}{\delta} $$ $$ \Rightarrow\left\Vert \Lambda\frac{x}{\delta}\right\Vert <\frac{\epsilon}{\delta} $$ $$ \Rightarrow\sup\left\Vert \Lambda\frac{x}{\delta}\right\Vert \leq\frac{\epsilon}{\delta} $$ $$ \Rightarrow\left\Vert \Lambda\right\Vert \leq\frac{\epsilon}{\delta} $$
Can you please clarify if there are mistakes in the above?
Can you also suggest alternative ways of deriving the above result?
In the last step above, we have used the definition of $\Lambda$ (Definition 5.3) which is given below. Here we are not using the condition that $\left\Vert{x}\right\Vert < \delta$ which would also imply that $\left\Vert\frac{x}{\delta}\right\Vert<1$ and not $\left\Vert\frac{x}{\delta}\right\Vert\leq1$ as needed by Definition 5.3
Also Definition 5.3 shows that $\Lambda$ is the smallest number such that the following inequality holds for every $x\in X$. Perhaps we can use this instead.
$$ \left\Vert \Lambda x\right\Vert \leq\left\Vert \Lambda\right\Vert \left\Vert x\right\Vert $$
Reference: Rudin's RCA $5.3$ Definition
Please let me know if any additional clarifications are needed. I could be overlooking something simple and hence happy to delete this question if that is the case.