A copy of $l_\infty$ in a infinite dimensional Banach space 
Let $E$ an infinite dimensional Banach space.
  
  
*
  
*Using the Hahn-Banach extension theorem, prove that there is a sequence $(y_n)\subset E$ and a decreasing sequence of closed
  subspaces $(M_n)$ such that $\sum\limits_{n=1}^\infty \Vert y_n \Vert < \infty$, $y_n \in M_{n-1}$ and $y_n \notin M_n$.
  
*Define $T:l_\infty \longrightarrow E$ by $T(x)=\sum\limits_{n=1}^\infty x_n y_n $ where $x=(x_n)\in l_\infty$.
  Prove that $T$ is linear and injective.

I tryied to use a countable subset of an algebric basis of E to construct such spaces in (1), but it didn't worked. In addition, assuming (1), I wasn't able to prove that T is injective. (I'm trying to prove that T is an isometry)
please, can anyone help me? (there is no need of an entire proof, just a sketch will help me a lot)
 A: Start with $M_0 = E$. Pick $y_1 \in E$, say with $\lVert y_1\rVert = 1$. By the Hahn-Banach theorem, there is an $f_1 \in E'$ with $\lVert f_1\rVert = 1$ and $f_1(y_1) = 1$. Let $M_1 = \ker f_1$.
After having found $y_k$ and $M_k$ for $1 \leqslant k \leqslant m$ with $\lVert y_k\rVert = 2^{1-k}$, $y_k \in M_{k-1}\setminus M_k$ and $\dim (E/M_k) = k$, you pick $y_{m+1} \in M_m$ with $\lVert y_{m+1}\rVert = 2^{-m}$ - that is possible since $M_m$ is infinite-dimensional - and use the Hahn-Banach theorem to find a suitable $M_{m+1}$ - pick $f_{m+1}\in E'$ with $\lVert f_{m+1}\rVert = 1$ and $f_{m+1}(y_{m+1}) = 2^{-m}$, then let $M_{m+1} = M_m \cap \ker f_{m+1}$. By the finiteness of the codimension of each $M_k$ in $E$, you can continue the construction.
To see that $T$ is injective, you need to see $\ker T = \{0\}$, so $x\neq 0 \implies T(x) \neq 0$. Pick $x\neq 0$ and let $k$ be the smallest index such that $x_k \neq 0$. Then
$$\sum_{n=k+1}^\infty x_ny_n = \lim_{m\to\infty} \sum_{n=k+1}^m x_n y_n \in M_k$$
since the finite sums belong to $M_k$ - $y_n \in M_{n-1} \subset M_k$ for $n > k$ - and $M_k$ is closed. But $y_k \notin M_k$, and therefore
$$T(x) = x_k y_k + \sum_{n=k+1}^\infty x_n y_n \notin M_k,$$
which implies $T(x)\neq 0$.
