Bounded and surjective map from counting $L^1$ to separable Banach space. 
Let $\mathscr{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbb{N}$. Suppose that $\{x_n\}_{1}^{\infty}$ is a countable dense subset of the unit ball of $\mathscr{X}$, and define $T:L^{1}(\mu)\rightarrow \mathscr{X}$ by $Tf = \sum_{1}^{\infty}f(n)x_n$.
a.) $T$ is bounded.
b.) $T$ is surjective.

Attempted proof a.) We have that $Tf = \sum_{1}^{\infty}f(n)x_n$, hence (from Henry) $$\lVert Tf\rVert  = \lVert \sum_{1}^{\infty}f(n)x_n\rVert \leq \sum_{1}^{\infty}|f(n)|\lVert x_n\rVert \leq \sum_{1}^{\infty}|f(n)| = \lVert f\rVert_{1}$$
I am not sure where the last part comes from I thought $\lVert f\rVert_{1} = \int |f|d\mu$.
Attempted proof b.) I need to show that if for all $y\in \mathscr{X}$ there exists an $x\in L^{1}(\mu)$ such that $T(x) = y$.
So, suppose $y\in\mathscr{X}$ and $x_n\rightarrow x$. Let $T_n\rightarrow T$ such that $$\lVert T_n x_n - Tx\rVert \rightarrow 0 \ \text{as} \ n\rightarrow \infty$$
Can I suppose now that there is a unique $y$ somewhere after this to show that that $T$ is surjective?
 A: Proof that $T$ is surjective.
(0) Obviously the $0$ vector of $X$ is in the image of $T.$
(1) For $x\in X$ and $\|x\|\ne 0,$  let $x'=x/\|x\| .$
(2) Let $j(1)$ be the least $j$ such that $0<\|x_j-x'\|<1/2.$ Let $f(j(1))=1.$
$$\text {(3)......................Let }\quad z_n=x'-\sum_{i=1}^{n}x_{j(i)}f(j(i)).$$ We have $0<\|z_1\|<1/2.$ Suppose  $0<\|z_n\|<2^{-n}.$  Let $j(n+1)$ be the least $j>j(n)$ such that $$0<\|x_j-(z_n/\|z_n\|)\|<1/2.$$( This is possible because $\{x_j\}_{j\in N}\backslash S$ is dense in the closed unit ball for any finite set $S.$)
Let $f(j(n+1))=\|z_{n+1}\|.$
(4) We have $$z_{n+1}    =z_n-x_{j(n+1)}f(j(n+1))=z_n-x_{j(n+1)}\|z_n\|=$$ $$=\|z_n\| \cdot \|(z_n/\|z_n\|)-x_{j(n+1)}\|.$$ $$\text {So }\quad 0<\|z_n\|<2^{-n}\implies 0<\|z_{n+1}\|<2^{-(n+1)}.$$ 
(5) Finally let $g(m)=0$ for $m\not \in \{j(n):n\in N\}$ and $g(j(n))=\|x\|f(j(n)).$ We  have $$\sum_{m\in N} | g(m)|=\sum _{n \in N}\|x\|f(j(n))\leq \sum_{n\in N} \|x\|2^{-n}=2\|x\|.$$ If $m'$ is the largest $j(i)$ not exceeding $m,$ then $m'\to \infty$ as $m\to \infty$ (because $j(n+1)>j(n)$),and for $m\geq j(1)$ we have $$\|x-\sum_{i\leq m} x_ig(i)\|=\|z_{m'}\|<2^{-m'}.$$ So $x=T(g).$
