For countable families $\{ x_n : n \in \mathbb{N}\}$ in a Banach (or Fréchet) space, summability and unconditional convergence are the same.
Let $\mathscr{F} = \{ F \subset \mathbb{N} : \operatorname{card}F < \infty\}$ the directed (by inclusion) set of finite subsets of $\mathbb{N}$, and for $F\in\mathscr{F}$, let $s_F = \sum\limits_{n\in F} x_n$.
The family is (by definition) summable if and only if the net $(s_F)_{F\in \mathscr{F}}$ converges (to $\lambda$, say).
Let the family be summable. Every bijection $\sigma \colon \mathbb{N}\to \mathbb{N}$ induces a map $\sigma^\ast \colon \mathbb{N}\to \mathscr{F}$ via $\sigma^\ast(n) = \{ \sigma(k) : 0 \leqslant k \leqslant n\}$, and $\{\sigma^\ast(n) : n \in \mathbb{N}\}$ is cofinal in $\mathscr{F}$, so $s^\sigma \colon n \mapsto s_{\sigma^\ast(n)}$ is a subnet of $(s_F)_{F\in \mathscr{F}}$ and hence also converges to $\lambda$. That is the unconditional convergence of $\sum\limits_{n=0}^\infty x_n$.
For the converse, we argue indirectly. Suppose the family is not summable. We have to show that it isn't unconditionally convergent.
Consider the sequence of partial sums
$$s_n = \sum_{k=0}^n x_k.$$
If that sequence doesn't converge, we're done, otherwise let $\mu$ be its limit.
That the family isn't summable to $\mu$ means that there is an $\varepsilon > 0$, such that for every $F_1 \in \mathscr{F}$, there is a $c(F_1)\in \mathscr{F}$ with $F_1 \subset c(F_1)$ and
$$\left\lVert\sum_{n\in c(F_1)} x_n - \mu \right\rVert \geqslant \varepsilon.$$
We can now construct a bijection $\sigma \colon \mathbb{N}\to\mathbb{N}$ such that
$$s^\sigma_n = \sum_{k=0}^n x_{\sigma(k)}$$
does not converge to $\mu$, which shows that $\{x_n\}$ is not unconditionally convergent.
Let $n_1 = \min \{n\in \mathbb{N} : \lVert s_n - \mu\rVert \leqslant \varepsilon/2\}$. Whenever $n_k$ is defined, we let $F_k = \{0,\dotsc,n_k\}$. For $k \leqslant n_1$, let $\sigma(k) = k$. Let $m_1 = \operatorname{card} c(F_1)$, and $\sigma(n_1+1),\dotsc,\sigma(m_1)$ the elements of $c(F_1)\setminus F_1$ in ascending order. Let $r_1 = \max c(F_1)$ and $n_2 = \min \{n > r_1 : \lVert s_n-\mu\rVert \leqslant \varepsilon/2\}$, and $\sigma(m_1+1),\dotsc,\sigma(n_2)$ the elements of $F_2\setminus c(F_1)$ in ascending order.
We continue in this way, when $n_k$ and $\sigma\lvert_{F_k}$ are determined, we let $m_k = \operatorname{card} c(F_k)$, $\sigma(n_k+1),\dotsc,\sigma(m_k)$ the elements of $c(F_k)\setminus F_k$ in ascending order, $r_k = \max c(F_k)$, $n_{k+1} = \min \{ n > r_k : \lVert s_n - \mu\rVert \leqslant \varepsilon/2\}$, and $\sigma(m_k+1),\dotsc,\sigma(n_{k+1})$ the elements of $F_{k+1}\setminus c(F_k)$ in ascending order.
The construction yields $n_k < m_k \leqslant r_k < n_{k+1}$ for all $k$, and
$$\left\lVert \sum_{k=0}^{m_k} x_{\sigma(k)} - \mu\right\rVert \geqslant \varepsilon,$$
so $(s^\sigma_n)$ does not converge to $\mu$ (it does not converge at all, in fact).
