I will try to give a more geometric explanation. First note that you can always scale $T$ such that it has norm one. Hence, the Banach-Mazure distance can be rewritten as:
$$
d(X,Y)=\inf\{||T^{-1}||: T\in GL(X,Y), ||T||=1\}
$$
Geometrically, $||T||=1$ means that $T(B_X)\subseteq B_Y$ and no enlargement of $T(B_X)$ will still fit inside $B_Y$ ($B_X$ and $B_Y$ represent the unit balls of the two spaces). On the other hand, we have that
$$
T^{-1}(B_Y)\subseteq||T^{-1}||B_X
$$
or equivalently
$$
B_Y\subseteq||T^{-1}||T(B_X)
$$
Threfore:
$$
T(B_X)\subseteq B_Y\subseteq||T^{-1}||T(B_X)
$$
Thus, geometrically, $||T^{-1}||$ represent the smallest amount by which you must increase $T(B_X)$ such that it contains $B_Y$. The Banach-Mazur distance represents the infimum of such enlargments, taken over all linear isomorphism that send $B_X$ inside $B_Y$.
For a perhaps a better intuitive understanding, take $B_X$ to be the unit sphere. Then for any isomorphism $T$, $T(B_X)$ is going to be an ellipsoid. For the banach-Mazur distance, you are looking for the "best fit" ellipsoid. That is, you are looking for the ellipsoid that fits inside $B_Y$ (touching the boundary), such that the enlargement required for this ellipsoid to contain $B_Y$ is as small as possible.
Yes, there is a similar notion for infinite dimensional Banach spaces, with the convention that when $X$ and $Y$ are not isomorphic, $d(X,Y)=\infty$.
Best Answer
For the beginning the theorem should be read as follows:
Proof. Consider $K:=\operatorname{Ball}_X(0,1)$ with the weak-$^*$ topology. By the Banach-Alaoglu theorem it is compact. By definition of the weak-$^*$ topology the map $$ J(x): K\to\mathbb{C}:f\mapsto f(x) $$ is continuous for each $x\in X$. So we have a well-defined map $$ J:X\to C(K):x\mapsto J(x) $$ Note that $$ \Vert J(x)\Vert=\sup\{|J(x)(f)|:f\in K\}=\sup\{|f(x)|:f\in \operatorname{Ball}_X(0,1)\}=\Vert x\Vert $$ for each $x\in X$. In the last step we use a corollary of the Hahn-Banach theorem. Thus $J$ is an isometric embedding.
Note that the corollary of the Hahn-Banach theorem does not require completeness, so its usage was valid in the proof above.