Continuous, bijective linear operator between Banach spaces has continuous inverse (Brezis) Corollary 2.7 from Brezis, Functional Analysis is stated below.

Let $E$ and $F$ be two Banach spaces and let $T$ be a continuous linear operator from $E$ to $F$ that is bijective, i.e., injective and surjective. Then $T^{-1}$ is also continuous (from F into E).
Proof. Property (7) and the assumption that $T$ is injective imply that if $x \in E$ is chosen so that $||Tx|| < c$, then $||x|| < 1$. By homogeneity, we find that 
  \begin{align*}
||x|| \leq \frac{1}{c} ||Tx|| \hspace{.5em} \forall x \in E
\end{align*}

Property (7) was that if $E$ and $F$ are two Banach spaces and $T$ is a continuous linear operator from $E$ into $F$ that is surjective, then there exists some constant $c > 0$ such that $T(B_E(0,1)) \supset B_F(0,c)$.
I'm having trouble understanding the line: "By homogeneity, we find that ..."
So I tried assuming that there exists an $x \in E$ such that
\begin{align*}
||x|| > \frac{1}{c} ||Tx||
\end{align*}
Then we could write $c||x|| > ||Tx|| \Rightarrow c > \frac{||Tx||}{||x||}$. But since $||x|| < 1$, we then have $c > ||Tx||$. But this doesn't seem to contradict anything, and I'm not sure why homogeneity applies here.
Any suggestions would be helpful!
 A: So, we know from the open mapping theorem,
$$(\exists c > 0)\quad ||Tx|| < c \implies ||x||<1.$$
It follows that if $||x|| \geq 1,$ then $||Tx|| \geq c$. 
For $x=0$ the statement is true (because then $||Tx||=||x||=0$), so let $x \neq 0$. 
$$||\frac{x}{||x||}|| \geq 1,$$
so $$||\frac{Tx}{||x||}|| \geq c,$$
from which the result follows.
A: Suppose $\|x\| > \frac{1}{c}\|Tx\|$. Then $x\ne 0$, so we can say that $\frac{\|Tx\|}{\|x\|} < c$, and in particular we can say that there exists $\epsilon>0$ such that $\frac{\|Tx\|}{\|x\|} = c-\epsilon$. Now, let
$$ y = \frac{c-\epsilon/2}{\|Tx\|}x. $$
Then
$$ \|y\| = (c-\epsilon/2)\frac{\|x\|}{\|Tx\|} = \frac{c-\epsilon/2}{c-\epsilon} > 1 $$
but
$$ \|Ty\| = \left\|T\left(\frac{c-\epsilon/2}{\|Tx\|}x\right)\right\| = \frac{c-\epsilon/2}{\|Tx\|}\|Tx\| = c-\epsilon/2 < c $$
so $y$ satisfies $\|Ty\| < c $ but $\|y\| > 1$, contradicting the assumption that $\|Ty\|<c\implies \|y\|<1$.

We can also argue this directly, as follows: suppose $x\ne 0$, so that $Tx\ne 0$ by injectivity. We know that for any $\alpha>0$, we have
$$\|x\| = \frac{\|\alpha x\|}{\alpha} = \frac{\|\alpha x\|\|Tx\|}{\alpha\|Tx\|} = \frac{\|\alpha x\|}{\|T(\alpha x)\|}\|Tx\|.$$ 
The ratio $k = \frac{\|\alpha x\|}{\|T(\alpha x)\|}$ is a constant independent of $\alpha$. Now, if we choose $\alpha$ small enough so that $\|T(\alpha x)\|<c$, then we would have $\|\alpha x\|<1$. Thus, for all $\alpha>0$, we have that
$$\|T(\alpha x)\|<c \implies \frac{\|\alpha x\|}{\|T(\alpha x)\|} < \frac{1}{\|T(\alpha x)\|} $$
and hence $k = \frac{\|\alpha x\|}{\|T(\alpha x)\|} < \frac{1}{\|T(\alpha x)\|}$ for all $\|T(\alpha x)\|<c$. This implies that $k\le\frac{1}{c}$, and so $\|x\| = k\|Tx\|\le \frac{1}{c}\|Tx\|$.
