Let $I \neq \emptyset$ set and $(x_i)_{i \in I}$ a family in a normed space $X$. We say that $(x_i)_{i \in I}$ is unconditionally Cauchy if $\forall \epsilon>0$ exists $F_{\epsilon} \subset I$ finite such that for all finite subset $F' \subset I$ with $F' \cap F_{\epsilon} = \emptyset$ we have:
$$||\sum_{i \in F'}x_i|| < \epsilon$$
I'm trying to show that if $(x_i)_{i \in I}$ is unconditionally cauchy then $A=\{ i \in I : x_i \neq0 \}$ is countable. I thought writing $A= \bigcup_{n \in \mathbb{N}}A_n $ where $A_n=\{ i \in I: ||x_i||>\frac{1}{n} \}$. Now the problem is fixed $n_0$ show that $A_{n_0}$ is countable.I tried to show by contradiction, but I did not succeed.
Claim: $A=\bigcup_n F_{1/n}$.
Proof: Suppose $x_i\neq 0$. Then $\lVert x_i\rVert>1/n$ for some $n$. If $i\notin F_{1/n}$ then choosing $F'=\{i\}$, we have $F'\cap F_{1/n}=\varnothing$ and hence by definition of $F_{1/n}$: $$ \left\lVert\sum_{j\in F'} x_j\right\rVert<\frac1n $$ but LHS is $\lVert x_i\rVert>\frac1n$, contradiction. QED.
Now each $F_{1/n}$ is finite, so $A$ is a countable union of finite, hence countable.