I wanted to ask some clarification on one of the proofs in Rudin's PMA, specifically Theorem 3.11 (c). It states that
In $\mathbb R^k$, every Cauchy sequence converges.
The proof for it is as follows:
Let $\{\mathbf x_n\}$ be a Cauchy sequence in $\mathbb R^k$. Define $E_N$ to be the set $\mathbf x_N, \mathbf x_{N + 1}, \ldots$. Then for some $N$, we have that the diameter of $E_N$ is less than 1. The range of $\{\mathbf x_n\}$ is the union of $E_N$ and the finite set $\{\mathbf x_1, \mathbf x_2, \ldots, \mathbf x_{N - 1}\}$. Then $\{\mathbf x_n\}$ is bounded. Since every bounded subset of $\mathbb R^k$ has compact closure in $\mathbb R^k$ by Theorem 2.41, then (c) follows from (b).
In the bolded part, I have no clue as to how we can just say that there exists some N. Is there some property of Cauchy sequences that I'm missing?
Since $\{x_n\}$ is Cauchy, by the definition of Cauchy sequence, exists $N$ s.t. $m,n>N\Rightarrow {\rm d}(x_m,x_n)<\frac{1}{2}$. We claim the set $E_{N+1}=\{x_{N+1}, x_{N+2},\cdots\}$ has diameter less than $1$. In fact take arbitrary $a,b\in E_{N+1}$. Say $a=x_m, b=x_n (m,n>N)$. Then ${\rm d}(a,b)={\rm d}(x_m,x_n)<\frac{1}{2}$ \begin{equation*} \Rightarrow {\rm diam}E_{N+1}:=\sup_{a,b\in E_{N+1}}{\rm d}(a,b)\le \frac{1}{2}<1. \end{equation*}