Theorem: Infinite-Dimensional Vector Subspace of P-Sequence Space contains Topologically Complemented Subspace Linearly Isomorphic to Whole Space

Let $\GF \in \set {\R, \C}$.
Let $p \in \hointr 1 \infty$.
Let $\struct {\map {\ell_p} \GF, \norm {\, \cdot \,}_p}$ be the $p$-sequence space.
Let $Y$ be an infinite-dimensional closed vector subspace of $\map {\ell_p} \GF$.
Then there exists a closed vector subspace $Z$ of $\map {\ell_p} \GF$ that is topologically complemented in $\map {\ell_p} \GF$ and linearly isomorphic to $\map {\ell_p} \GF$.

Proof:
Let $\sequence {e_n}_{n \mathop \in \N}$ be the Schauder basis for $\map {\ell_p} \GF$ given by P-Sequence Space admits Schauder Basis.
Let $\sequence {e_n^\ast}_{n \mathop \in \N}$ be the coordinate functionals associated with $\sequence {e_n}_{n \mathop \in \N}$.
We show that, for each $n \in \N$, we can find $x_n \in Y$ with $\norm {x_n}_Y = 1$ and:
:$\map {e_k^\ast} {x_n} = 0$ for $1 \le k \le n$.
{{AimForCont}} that this is not true. 
Then there exists $N \in \N$ such that whenever:
:$\map {e_k^\ast} x = 0$ for $1 \le k \le N$
we have $x = {\mathbf 0}_X$.
Define the projection $P_N : X \to X$ by:
:$\ds \map {P_N} x = \sum_{j \mathop = 1}^N \map {e_j^\ast} x e_j$ for each $x \in X$.
From Coordinate Functionals Associated with Schauder Basis of Banach Space are Bounded: Corollary 1, $S_N$ is bounded.
We show that $P_N$ is injective. 
Suppose that for some $x \in X$:
:$\ds \sum_{j \mathop = 1}^N \map {e_j^\ast} x e_j = {\mathbf 0}_X$
From the definition of a Schauder basis, we have:
:$\map {e_k^\ast} x = 0$ for $1 \le k \le N$
and hence $x = {\mathbf 0}_X$.
Hence $\map \ker {P_N} = \set { {\mathbf 0}_X}$.
From Linear Transformation is Injective iff Kernel Contains Only Zero, $P_N$ is injective. 
Hence the restriction $P_N \restriction_Y : Y \to P_N \sqbrk Y$ is a bijective bounded linear transformation.
From the Banach Isomorphism Theorem, $P_N \restriction_Y : Y \to P_N \sqbrk Y$ is a linear isomorphism.
From Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space: Corollary, we have:
:$\map \dim {P_N \sqbrk Y} = \dim Y$
On the other hand, we have:
:$P_N \sqbrk Y = \map \span {\set {e_1, e_2, \ldots, e_N} }$
and hence:
:$\map \dim {P_N \sqbrk Y} < \infty$
This contradicts that $Y$ is infinite-dimensional.
Hence for each $n \in \N$, there exists $x_n \in Y$ with $\norm {x_n}_Y = 1$ and:
:$\map {e_k^\ast} {x_n} = 0$ for $1 \le k \le n$.
Fix such a sequence $\sequence {x_n}_{n \mathop \in \N}$.
Now, for each $k \in \N$ we have that:
:$\map {e_k^\ast} {x_n} = 0$ for $k > n$. 
Hence:
:$\map {x_n} k \to 0$ as $n \to \infty$
for each $k \in \N$. 
From Normalized Sequence in P-Sequence Space which Vanishes Pointwise has Basic Subsequence Equivalent to Schauder Basis, $\sequence {x_n}_{n \mathop \in \N}$ has a subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$ that is basic, equivalent to $\sequence {e_n}_{n \mathop \in \N}$ and complemented in $\map {\ell_p} \GF$.
Let $Z = \sqbrk {x_{n_k} }_{k \mathop \in \N}$ be the closed linear span of $\sequence {x_{n_k} }_{k \mathop \in \N}$.
Then $Z$ is topologically complemented in $\map {\ell_p} \GF$.
From Basic Sequences in Banach Spaces are Equivalent iff Map between is Linear Isomorphism, there exists a linear isomorphism $T : \sqbrk {x_{n_k} }_{k \mathop \in \N} \to \sqbrk {e_n}_{n \mathop \in \N}$.
That is, there exists a linear isomorphism $T : Z \to \map {\ell_p} \GF$.
Hence $Z$ is linearly isomorphic to $\map {\ell_p} \GF$.
{{qed}}