I'm trying to understand that proof with little success. Particularly I don't understand why "S has x_0 as a limit point, and S has no other limit point in R^k". Please help
The conditions [itex]|x_n-x_0|<1/n[/itex] imply that [itex](x_n)_n[/itex] converges to [itex]x_0[/itex]. Thus [itex]x_0[/itex] is a limit point of the [itex](x_n)_n[/itex] (which is the set S). Since limits of sequences are unique in [itex]\mathbb{R}^k[/itex], the limit point [itex]x_0[/itex] is unique.
Alright, but can you explain it with basic topology, like neighbourhoods and the definition of a limit point? I don't see why any other interior point of S couldn't be a also a limit point in that case. Rudin says that it's because it would violate the triangular inequality theorem, but still it's not clear enough...
If y is a limit point of [itex](x_n)_n[/itex], then there exists a subsequence [itex](x_{k_n})_n[/itex] that converges to y. But since the original sequence [itex](x_n)_n[/itex] converges to x, the subsequence must also converge to x.
So [itex](x_{k_n})_n[/itex] converges to both x and y. Thus x must equal y.