I'm reading the proof of open mapping theorem in Brezis' book of Functional Analysis.
Let $T: E \to F$ be a linear continuous function between Banach spaces. We denote the unit open balls by $\mathbb B_E := \{x\in E \mid \|x\| <1\}$ and $\mathbb B_F := \{y\in F \mid \|y\| <1\}$. If $T$ is surjective, then there is $c>0$ such that $$c\mathbb B_F \subseteq T[\mathbb B_E].$$
The author first derives that there is $c>0$ such that $2c\mathbb B_F \subseteq \overline{T[\mathbb B_E]}$. Then he goes on to prove $c\mathbb B_F \subseteq T[\mathbb B_E]$ as follows. Choose any $y \in F$ with $\|y\|<c$. The aim is to find some $x \in E$ such that $$ \|x\|<1 \quad \text { and } \quad T x=y . $$
We have $2c\mathbb B_F \subseteq \overline{T[\mathbb B_E]}$, so $\forall \varepsilon>0, \exists z \in E$ such that $\|z\|<\frac{1}{2}$ and $\|y-T z\|<\varepsilon$. Choosing $\varepsilon=c / 2$, we find some $z_{1} \in E$ such that $\left\|z_{1}\right\|<\frac{1}{2}$ and $\left\|y-T z_{1}\right\|<\frac{c}{2}$. By the same construction applied to $y-T z_{1}$ (instead of $y$ ) with $\varepsilon=c / 4$ we find some $z_{2} \in E$ such that $\left\|z_{2}\right\|<\frac{1}{4}$ and $\left\|\left(y-T z_{1}\right)-T z_{2}\right\|<\frac{c}{4}$. Proceeding similarly, by induction we obtain a sequence $\left(z_{n}\right)$ such that $$ \left\|z_{n}\right\|<\frac{1}{2^{n}} \text { and }\left\|y-T\left(z_{1}+z_{2}+\cdots+z_{n}\right)\right\|<\frac{c}{2^{n}} \quad \forall n . $$ It follows that the sequence $x_{n}=z_{1}+z_{2}+\cdots+z_{n}$ is a Cauchy sequence. Let $x_{n} \rightarrow x$ with, clearly, $\|x\|<1$ and $y=T x$ (since $T$ is continuous).
We have $$\|x_n\| \le \sum_{k=1}^n \|z_k\| < \sum_{k=1}^n \frac{1}{2^k} = 1 -\frac{1}{2^n} <1.$$ Then $|x| = \lim |x_n| \le 1$. Could you explain how the author obtains $\|x\| < 1$?
By triangle inequality, we have $$\|x\| = \left \| \sum_{n=1}^\infty z_n \right \| \le \sum_{n=1}^\infty \left \| z_n \right \| < \sum_{n=1}^\infty \frac{1}{2^n} = 1.$$