To prove that the space $C_0$ of the sequences x=($x_1,x_2$,...) for which $\displaystyle \lim_{i=\infty} x_i=0$ with the norm $\|x\|=\max_{1 \leq i < \infty} |x_i|$ does not have a two-dimensional subspace isometric to $E^2$, suppose that the space $C_0$ has a two-dimensional subspace isometric to $E^2$ then there would exist two points $x, y$ in $C_0$ for which $\|tx+y\|=(t^2+1)^{1/2}$ for every real $t$. Then, for every t there would exist at least one integer i for which $(t x_i + y_i)^2 = t^2+1$. This equation has at most two solutions for t, given $x_i$ and $y_i$, and so we are led to a contradiction since t may take any value of the nondenumerable set of the real line.
Q1 I don't know why if $c_0$ has a two-dimensional subspace isometric to E^2 then there would exist two points $x, y$ in $C_0$ for which $\|tx+y\|=(t^2+1)^{1/2}$ for every real $t$.
Q2 if we have two isometric spaces then what is the relation between their bases ?
Q3 why we have in final a contradiction
Suppose $V$ is such a subspace. Then there is a surjective isometry $T\colon E^2\to V$. Let $x = T((1,0))$ and $y=T((0,1))$. Then $$ \|tx+y\|_{c_0} = \|(t,1)\|_{E^2} = \sqrt{t^2+1} $$
The relation between basis is that the image of a basis under an invertible linear map is also a basis.
The contradiction is because for each of uncountably many $t$ we must have a suitable integer $i$, and the same $i$ can only work for two values of $t$.