The last part of this proof gives me some hints. For , find a such that . Define a mapping where is defined as (67) in P229 of this book satisfying on . . It can be seen that is a linear transformation from into , so it is continuous and in turn makes continuous. Therefore, the open set of produces by continuity of a neighborhood in containing such that for all . Note that so far may not lie in , so next we prove that it really is contained in . For any in , let , then , so , which just shows , or, , as desired. Note that this means is only open in , but not open in because is not necessarily open in . However, if we set the codomain to be the rank space assuming subspace topology, any linear transformation is an open map. Is that right?

I don't know if the above proof is what the author originally intended. It seems not and there should be a simpler proof since "it is clear". oops... Rudin's book is going to drive me mad , should I continue to use his books to study real and complex analysis and functional analysis in future?