It looks very simple, but I couldn't check this.
Let $(X, \|\cdot\|)$ be a vector space of dimension $n$ (over $\mathbb{R}$ or $\mathbb{C}$), where $\| \cdot \|$ is a norm on $X$.
Let $\{e_1, \cdots, e_n\}$ be an algebraic basis of $X$ with $\|e_i\| =1$ for all $1\leq i \leq n$.
For $x \in X$, write $x = a_1 e_1 + \cdots + a_n e_n$, where $a_1, \cdots a_n$ are scalars.
If $\|x\| \leq 1$, then is it true that $|a_1|, \cdots, |a_n| \leq 1$?
No, it's not true. Here's a counterexample in $\mathbb{R}^2$ with the usual Euclidean norm.
The green circle has radius $1$. The basis vectors $e_1$ and $e_2$ are the black ones. The red vector $x$ obviously has $\|x\| <1$. But also $x = \tfrac32e_1 + \tfrac32e_2$, so $a_1=a_2 = \tfrac32$.