Let $n \in \mathbb N$ and $(x^k)$ a sequence in $\mathbb K^n$ with $x^k= (x_1^k,..., x_n^k$) and it is given that $|x_j^k-x_j^{k+1}|\le 2^{-k}$ for all $j= 1,...,n$ and all $k$.
I want to show that $x^k$ is a Cauchy sequence, however I am stuck. My idea is to use that if all subsequences of $x^k$ are Cauchy therefore $x^k$ must be Cauchy combined with the fact that you can $\forall_{j=1, \ldots, n}\left|z_{j}\right| \leq \underbrace{|z|}_{\sqrt{\left|z_{1}\right|^{2}+\cdots+\left|z_{n}\right|^{2}}} \leq\left|z_{1}\right|+\cdots+\left|z_{n}\right| .$.
Notice that $$\|\mathbf{x}\|_\infty=\max_{1\leq j\leq n}|x_j|\leq\|\mathbf{x}\|_2=\sqrt{\sum^n_{j=1}|x_j|^2}\leq n\|\mathbf{x}\|_\infty$$ where $\mathbf{x}=(x_1,\ldots,x_n)$. So, it is enough to show that the sequence $\mathbf{x}^k$ is Cauchy in $(\mathbb{K}^d,\|\;\|_\infty)$. The conditions in the posting mean that $\|\mathbf{x}^{k+1}-\mathbf{x}^k\|_\infty\leq 2^{-k}$. So, for $m<k$ $$\begin{align} \|\mathbf{x}^k-\mathbf{x}^m\|_\infty&\leq\|\mathbf{x}^{k}-\mathbf{x}^{k-1}\|_\infty+\ldots+\|\mathbf{x}^{m+1}-\mathbf{x}^k\|_\infty\leq 2^{-(k-1)}+\ldots 2^{-m}\\ &\leq 2^{-m}(1+\ldots 2^{-(k-m+1)})\leq 1\cdot 2^{-m}\xrightarrow{m\rightarrow\infty}0 \end{align}$$
Thus, $\{\mathbf{x}^k:k\in\mathbb{N}\}$ is a Cauchy sequence.