Theorem: 1-Sequence Space Continuously Surjects into Separable Banach Space

Let $\GF \in \set {\R, \C}$.
Let $\struct {\map {\ell_1} \GF, \norm {\, \cdot \,}_1}$ be the $1$-sequence space.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a separable Banach space over $\GF$.
Then there exists a bounded linear surjection $Q : \map {\ell_1} \GF \to X$.

Proof:
Let $B_X$ be the open unit ball in $\struct {X, \norm {\, \cdot \,}_X}$. 
Let $B_X^-$ be the closed unit ball in $\struct {X, \norm {\, \cdot \,}_X}$.
Let $B_{\ell_1}$ be the open unit ball in $\struct {\map {\ell_1} \GF, \norm {\, \cdot \,}_1}$. 
Let $B_{\ell_1}^-$ be the closed unit ball in $\struct {\map {\ell_1} \GF, \norm {\, \cdot \,}_1}$.
By Subspace of Separable Metric Space is Separable, $\struct {B_X^-, \norm {\, \cdot \,}_X}$ is separable.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a countable set everywhere dense in $B_X^-$.
For each $\xi \in \map {\ell_1} \GF$, we have:
{{begin-eqn}}
{{eqn	| l = \sum_{n \mathop = 1}^\infty \norm {\map \xi n x_n}_X
		| r = \sum_{n \mathop = 1}^\infty \norm {x_n} \cmod {\map \xi n}
		| c = {{NormAxiomVector|2}}
}}
{{eqn	| o = \le
		| r = \sum_{n \mathop = 1}^\infty \cmod {\map \xi n}
}}
{{eqn	| o = <
		| r = \infty
		| c = {{Defof|p-Sequence Space}}
}}
{{end-eqn}}
From Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach, we have that:
:$\ds \sum_{n \mathop = 1}^\infty \map \xi n x_n$ converges for each $\xi \in \map {\ell_1} \GF$.
We can therefore define $Q : \map {\ell_1} \GF \to X$ by:
:$\ds \map Q \xi = \sum_{n \mathop = 1}^\infty \map \xi n x_n$ 
for each $\xi \in \map {\ell_1} \GF$.
We first show that $Q$ is a bounded linear transformation.
Let $\lambda \in \GF$ and $\xi_1, \xi_2 \in \map {\ell_1} \GF$.
We have:
{{begin-eqn}}
{{eqn	| l = \map Q {\xi_1 + \lambda \xi_2}
		| r = \sum_{n \mathop = 1}^\infty \map {\paren {\xi_1 + \lambda \xi_2} } n x_n
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \paren {\map {\xi_1} n + \lambda \map {\xi_2} n} x_n
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map {\xi_1} n x_n + \lambda \sum_{n \mathop = 1}^\infty \map {\xi_2} n
		| c = Linear Combination of Convergent Series: Normed Vector Space
}}
{{eqn	| r = \map Q {\xi_1} + \lambda \map Q {\xi_2}
}}
{{end-eqn}}
For each $\xi \in \map {\ell_1} \GF$:
{{begin-eqn}}
{{eqn	| l = \norm {\map Q \xi}_X
		| r = \norm {\sum_{n \mathop = 1}^\infty \map \xi n x_n}_X
}}
{{eqn	| o = \le
		| r = \sum_{n \mathop = 1}^\infty \cmod {\map \xi n} \norm {x_n}_X
		| c = {{NormAxiomVector|2}}, {{NormAxiomVector|3}}
}}
{{eqn	| o = \le
		| r = \sum_{n \mathop = 1}^\infty \cmod {\map \xi n}
}}
{{eqn	| r = \norm \xi_1
}}
{{end-eqn}}
We move towards showing that $B_X \subseteq Q \sqbrk {B_{\ell_1} }$.
Let $\sequence {e_n}_{n \mathop \in \N}$ be the Schauder basis for $\map {\ell_1} \GF$ determined in P-Sequence Space admits Schauder Basis.
Note that $\norm {e_n}_1 = 1$ for each $n \in \N$.
Hence since $\map Q {e_n} = x_n$ for each $n \in \N$, we have $\set {x_n : n \in \N} \subseteq Q \sqbrk {B_{\ell_1}^-}$.
Since $\sequence {x_n}_{n \mathop \in \N}$ is everywhere dense in $B_X^-$, we have $B_X^- \subseteq \map \cl {Q \sqbrk {B_{\ell_1}^-} }$ from Set Closure Preserves Set Inclusion.
Let $0 < \epsilon < 1$. 
Let $x \in B_X^-$.
Then there exists $\xi_1 \in B_{\ell_1}^-$ such that $\norm {x - Q \xi_1}_X < \epsilon$.
Then we have, from {{NormAxiomVector|3}}:
:$\ds \norm {\frac {x - Q \xi_1} \epsilon}_X < 1$ 
Hence there exists $\xi_2' \in B_{\ell_1}^-$ such that:
:$\ds \norm {\frac 1 \epsilon \paren {x - Q \xi_1} - Q \xi_2'}_X < \epsilon$
We then have:
:$\ds \norm {x - \paren {Q \xi_1 + \map Q {\epsilon \xi_2'} } }_X < \epsilon^2$
Hence setting $\xi_2 = \epsilon \xi_2'$, we have:
:$\ds \norm {x - \map Q {\xi_1 + \xi_2} }_X < \epsilon^2$
and:
:$\norm {\epsilon \xi_2'} \le \epsilon$
Suppose that we have found $\xi_1, \ldots, \xi_n \in B_{\ell_1}^-$ with $\norm {\xi_n}_1 \le \epsilon^{n - 1}$ and $\norm {x - \map Q {\xi_1 + \ldots + \xi_n} }_X < \epsilon^n$.
Then we have:
:$\ds \norm {\frac 1 {\epsilon^n} \paren {x - \map Q {\xi_1 + \ldots + \xi_n} } }_X < 1$ 
Hence there exists $\xi_{n + 1}' \in B_{\ell_1}^-$ such that:
:$\ds \norm {\frac 1 {\epsilon^n} \paren {x - \map Q {\xi_1 + \ldots + \xi_n} } - Q \xi_{n + 1}'}_X < \epsilon$
Hence, setting $\xi_{n + 1} = \epsilon^n \xi_{n + 1}'$ we have:
:$\ds \norm {\paren {x - \map Q {\xi_1 + \ldots + \xi_n} } - Q \xi_{n + 1} }_X < \epsilon^{n + 1}$
with $\norm {\xi_{n + 1} }_1 \le \epsilon^n$ and hence the construction continues.
We then have:
:$\ds \sum_{n \mathop = 1}^\infty \norm {\xi_n}_1 \le \sum_{n \mathop = 1}^\infty \epsilon^{n - 1}$
From Sum of Infinite Geometric Sequence, we have:
:$\ds \sum_{n \mathop = 1}^\infty \epsilon^{n - 1} = \frac 1 {1 - \epsilon}$
Hence from Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach:
:$\ds \sum_{n \mathop = 1}^\infty \xi_n$ converges to $\xi \in B_{\ell_1}^-$.
We have:
:$\ds \norm {x - \map Q {\sum_{n \mathop = 1}^N \xi_n} }_X < \epsilon^N$
Taking $N \to \infty$, we have:
:$\ds \norm {x - \map Q {\sum_{n \mathop = 1}^N \xi_n} }_X \to 0$
while:
:$\ds \map Q {\sum_{n \mathop = 1}^N \xi_n} \to Q \xi$
from Continuous Mappings preserve Convergent Sequences.
Hence we have $Q \xi = x$.
Hence we have shown that for each $x \in B_X^-$ and $\epsilon > 0$, there exists $\xi \in \map {\ell_1} \GF$ such that $\norm \xi_1 < \paren {1 - \epsilon}^{-1}$ and $Q \xi = x$.  
Now let $u \in B_X$.
Then there exists $\delta < 1$ such that $\norm u_X < \delta$.
We then have:
:$\ds \frac u \delta \in B_X$.
Take:
:$\ds \epsilon < 1 - \delta$
so that:
:$\ds \frac 1 {1 - \epsilon} < \frac 1 \delta$ 
By what we have shown, we can find $\xi \in \map {\ell_1} \GF$ with:
:$\ds \norm \xi_1 < \frac 1 {1 - \epsilon}$ 
such that:
:$\ds Q \xi = \frac u \delta$
Then, we have:
:$\map Q {\delta \xi} = u$
Further:
:$\ds \norm {\delta \xi}_1 < \frac \delta {1 - \epsilon} < 1$
so that $\delta \xi \in B_{\ell_1}$. 
We have therefore show that $B_X \subseteq Q \sqbrk {\ell_1}$. 
Hence:
{{begin-eqn}}
{{eqn	| l = X 
		| r = \bigcup_{\delta > 0} \delta B_X
}}
{{eqn	| o = \subseteq
		| r = \bigcup_{\delta > 0} \delta Q \sqbrk {B_{\ell_1} }
}}
{{eqn	| r = \bigcup_{\delta > 0} Q \sqbrk {\delta B_{\ell_1} }
		| c = {{Defof|Linear Transformation}}
}}
{{eqn	| r = Q \sqbrk {\bigcup_{\delta > 0} \delta B_{\ell_1} }
		| c = Image of Union under Mapping: General Result
}}
{{eqn	| r = Q \sqbrk {\map {\ell_1} \GF}
}}
{{eqn	| o = \subseteq 
		| r = X
}}
{{end-eqn}}
Hence $Q \sqbrk {\map {\ell_1} \GF} = X$.
Hence $Q$ is surjective.
We have also established that $Q$ is a bounded linear transformation.
{{qed}}