Example to show that unconditional convergence does not imply absolute convergence in infinite-dimensional normed spaces. I tried to show that unconditional convergence does not imply absolute convergence in infinite-dimensional normed spaces using a direct proof, but unfortunately I did not succeed. The definition of unconditional convergence I am using is the following:

A series $\sum_n x_n$ is called unconditionally convergent if the series $\sum_n x_{\pi(n)}$ converges for all permutations $\pi : \mathbb{N} \to \mathbb{N}$.

I considered the sequence $(x_n)_{n \in \mathbb{N}} \subset c_0 (\mathbb{N})$, where $x_n = \frac{\delta_n}{n}$ and $\delta_n$ denoting the standard unit vector. It is clear that $\sum_n \| x_n \|_{\infty}$ does not convergence, but I have trouble showing that the series $\sum_n x_n$ converges unconditionally. 
Any help is appreciated. Thanks in advance.

Denote by $y = (y^n)_{n \in \mathbb{N}}$ the sequence $y = (1, \frac{1}{2}, \frac{1}{3}, \dots)$ (we will use upper indices for the terms of an element in $\ell^{\infty}(\mathbb{N})$ in order to not confuse ourselves when considered sequences of elements in $\ell^{\infty}(\mathbb{N})$). Then we have
$$ \left( y - \sum_{k=1}^n x_{\pi(k)} \right)^i = \left( y - \sum_{k=1}^n \frac{\delta_{\pi(k)}}{k} \right)^i = \begin{cases} 0 & i \in \{ \pi(1), \dots, \pi(k) \}, \\ \frac{1}{i} & \textrm{otherwise}.  \end{cases} $$
Given $n > 0$, choose $N_0$ such that $\{ 1, \dots, n \} \subseteq \{\pi(1), \dots, \pi(N_0) \}$ (any $N_0 > \max \{ \pi^{-1}(1), \dots, \pi^{-1}(n) \}$ will do). Then, for all $N \geq N_0$, we will still have
$\{ 1, \dots, n \} \subseteq \{\pi(1), \dots, \pi(N) \}$ and so
$$ \left| \left| y - \sum_{k=1}^N x_{\pi(k)} \right| \right|_{\infty} = \sup_{i \in \mathbb{N}} \left|  \left( y - \sum_{k=1}^N x_{\pi(k)} \right)^{i} \right| = \sup_{i \in \mathbb{N} \setminus \{ \pi(1), \dots, \pi(N) \}} \left| \frac{1}{i} \right| \leq \frac{1}{n+1}. $$
