There are two (Why?) in the proof, i think they are trivial but, I don't see why they are true.
If anyone can show me some proof of these it would be very helpful.
Definition: Let $X,Y$ two normed vector spaces, and we define the Banach-Mazur distance between $X,Y$ as $$ d(X,Y)= \begin{cases} \inf \{ \|T\|\|T^{-1}\|:T\in GL(X,Y) \}, \quad X \sim Y\\
+\infty, \quad X \not\sim Y \end{cases} ,$$
where $GL(X,Y)$ is the space of all linear isomorphisms and $X \sim Y$ means $X$ isomorphic to $Y$ .
An equivalent definition is given by the following proposition:
**Proposition:**Let $X,Y$ two isomorphic normed vector spaces, the Banach-Mazur distance between $X,Y$ is given by:
$$ d(X,Y)=\inf\{d>0,|\exists T \in GL(X,Y) : \quad B_Y \subset T(B_X) \subset d B_Y \}$$
Where
$$ B_X = \{ x \in X : \|x\| \leq1 \} \quad \text{and} \quad B_Y = \{ y \in Y : \|y\| \leq1 \} $$
Proof: Suppose $d(X,Y)<d<+\infty$. From the definition there exists a linear isomorphism $T:X \rightarrow Y$ s.t.
$$ \|T\|\|T^{-1}\|<d$$
From the definition of a norm of an operator we have:
$a)$ For every $x \in B_X$ we have $\|Tx\|_Y\leq \|T\|\|x\|_X \leq \|T\|$ so,
$$T(B_X) \subset\|T\|B_Y \quad \text{(Why?)}$$
$b)$ For every $y \in B_X$ we have $\|T^{-1}y\|_X \leq \|T^{-1}\|\|y\|_Y \leq \|T^{-1}\|$ so, $$T^{-1}(B_Y) \subset\|T^{-1}\|B_X \quad \text{(Why?)}$$ Or equivalently, $$B_Y \subset \|T^{-1}\|T(B_X).$$ Lets set $S=\|T^{-1}\|T$, then, from $(a)$ we have $$S(B_X) \subset \|T\|\|T^{-1}\|B_Y $$ and from $(b)$ we have $$ B_Y \subset S(B_X)$$ So, there exist a $S:X \rightarrow Y$ s.t. $$ B_Y \subset S(B_X)\subset d B_Y.$$ From the other hand, if $ B_Y \subset S(B_X)\subset d B_Y$ for a $S:X \rightarrow Y$, Then $$\|S\| \leq d \quad \text{and} \quad \|S^{-1}\| \leq 1$$ So $$d(X,Y) \leq \|S\|\|S^{-1}\| \leq 1 $$
If $y\in T(B_X)$, then there exists an $x\in B_X$ for which $y=Tx$. By the previous line, we know that $\lVert Tx\rVert_Y\leqslant\lVert T\rVert$ hence $y$ belongs to $\lVert T\rVert B_Y=\{\lVert T\rVert z,z\in B_X\}$.
For the second question, this is exactly the same argument with $X$ replaced by $Y$, $Y$ by $X$ and $T$ by $T^{-1}$.