If $x=(x_n), s_n=\sum_{k=1}^n x_k$ and $\sup_n |s_n| =1$ then $\sup_n|x_n|=2$ In this case, $x\subseteq\mathbb{C}$. I can visualize the problem graphically, if I have a complex number $z$ inside the ball with radius 1, centered at the origin, I can add a lot of other complex numbers without leaving the ball, all of them with a norm smaller than 2. The only cases where I'll get a complex $w$ with norm 2, is for example $z=1+i$ and $w=-2-2i$. However I haven't found a way to write down this idea. I think there's a better way to prove it but I can't find it.

The statement is false: consider $x_1=1$ and $x_n=0$ for $n\ge 2$. Then $s_n=1$ for every $n$ so $\sup_{n}|s_n|=1$. But clearly, $\sup_n|x_n|=1<2$.
What can be shown: Since $x_n=s_n-s_{n-1}$, the Triangular Inequality $|x_n|=|s_n-s_{n-1}|\le |s_n|+|s_n-1|$ implies that $\sup_n|x_n|\le 2$. This is a formula of your geometric intuition.
