I am trying to understand the answer given in Brezis' book for this particular exercise:
Let $E$ and $F$ be two Banach spaces with norms $\|\ \|_E$ and $\|\ \|_F$. Let $T\in\mathcal L(E,F)$ be such that $R(T)$ is closed and $\dim N(T)<\infty$. Let $|\ |$ denote another norm on $E$ that is weaker that $\|\ \|_E$, i.e., $|x|\leqslant M\|x\|_E\ \forall x\in E$.
Prove that there exists a constant $C$ such that $$\|x|_E\leqslant C\left(\|Tx\|_F+|x|\right)\quad\forall x\in E.$$ Hint: Argue by contradiction
In the back of the book there is an answer to the exercise and I am trying to fill in the details, however there is a very important step here
Without loss of generality we may assume that $T$ is surjective (otherwise, replace $E$ by $R(T)$). Assume by contradiction that there is a sequence $(x_n)$ in $E$ such that $$ \|x_n\|_E=1\quad\text{ and } \|Tx_n\|_F+|x_n|<1/n.$$ By the open mapping theorem there is a constant $c>0$ such that $T(B_E)\supset cB_F$. Since $\|Tx_n\|_F<1/n$, there exists some $y_n\in E$ such that $$Tx_n = Ty_n \quad \text{ and } \|y_n\|_E < 1/nc. $$ Write $x_n+y_n+z_n$ with $z_n\in N(T)$, $\|y_n\|_E\to 0$, and $\|z_n\|_E\to1$. On the other hand, $|x_n|<1/n$; hence $|z_n|<(1/n)+|y_n\leqslant (1/n)+M\|y_n\|_E$, and consequently $|z_n|\to0$. This is impossible, since the norms $\|\ \|_E$ and $|\ |$ are equivalent on the finite-dimensional space $N(T)$.
which I do not understand.
Why can we assume $\|y_n\|_E \leq \frac{1}{nc}$? The open mapping provides us with the inequality $\|y_n\|_E \leq \frac{1}{c}$ since $cB_F \subseteq T(B_E)$, but without the 'n' in the denominator we can no longer assume that $y_n \to 0$ and we need that to get the contradiction.
You have $\|Tx_n\|_F<1/n$, so $nTx_n\in B_F\subset \frac1c\,T(B_E)$. Then there exists $y_n'\in B_E$ with $nTx_n=\frac1c\,Ty_n'$. Let $$y_n=\frac1{nc}\,y_n'.$$ Thus $Tx_n=Ty_n$, and $$ \|y_n\|_E=\frac1{nc}\,\|y_n'\|_E\leq\frac1{nc}. $$