= Solution
Here “subspace” is understood to mean a closed <closed linear subspace>. Closedness matters: the <finitely supported sequences> $c_{00}$ are an infinite-dimensional algebraic subspace of $\ell^p$, but contain no infinite-dimensional complete subspace. Indeed, any such subspace would be the countable union of its intersections with the finite-dimensional coordinate spaces, contrary to the <Baire category theorem>. We prove the assertion with the intended closedness convention, for $1\le p<\infty$.
Write $E=\ell^p$, the <l-p sequence space>. Choose positive numbers $\delta_n$ with $\sum_n2\delta_n<1$. Starting with $m_0=0$, choose $x_n\in X$ with $\|x_n\|_p=1$ and its first $m_{n-1}$ coordinates zero. This is possible because the kernel of finitely many coordinate <linear functionals> on an infinite-dimensional <vector space> is still infinite-dimensional. Choose $m_n>m_{n-1}$ so that the part of $x_n$ after coordinate $m_n$ has norm below $\delta_n$. Let $v_n$ be its restriction to $I_n=(m_{n-1},m_n]$, and put $u_n=v_n/\|v_n\|_p$. The <triangle inequality> gives
$$
\|x_n-u_n\|_p\le\|x_n-v_n\|_p+|1-\|v_n\|_p|<2\delta_n.
$$
The vectors $u_n$ form a normalized <block basic sequence>. Their supports are disjoint, so
$$
\left\|\sum_n a_nu_n\right\|_p^p=\sum_n|a_n|^p.
$$
Consequently $U=\overline{\operatorname{span}}\{u_n\}$ is isometric to the <l-p sequence space>.
For each block choose a norm-one <linear functional> $\phi_n$ supported on $I_n$ with $\phi_n(u_n)=1$. The <Hahn-Banach theorem> supplies such a functional, including when $p=1$. Define
$$
Qz=\sum_n\phi_n(z)u_n.
$$
The series converges in the <l-p sequence space>, since
$$
\|Qz\|_p^p=\sum_n|\phi_n(z)|^p\le\sum_n\|z|_{I_n}\|_p^p\le\|z\|_p^p.
$$
Also $Qu_n=u_n$, so $Q$ is a bounded projection onto $U$. Now define the <bounded linear operator>
$$
A=I+\sum_n\phi_n(\cdot)(x_n-u_n).
$$
This series converges in <operator norm> and $\|A-I\|\le\sum_n2\delta_n<1$. The <Neumann series> therefore makes $A$ invertible. Disjoint support gives $Au_n=x_n$. Thus $Y=AU=\overline{\operatorname{span}}\{x_n\}$ lies in the closed subspace $X$, is isomorphic to the <l-p sequence space>, and is the range of the bounded projection $AQA^{-1}$ on the whole space $E$. This proves \b[the required copy of $\ell^p$ is complemented in the ambient $\ell^p$], not just in $X$.
Now let $Z$ be an infinite-dimensional <complemented subspace> of $E$. The construction supplies $Y\subset Z$ with $Y\cong E$ and a bounded projection $R:E\to Y$. Its restriction to $Z$ gives the topological <direct sum>
$$
Z=Y\oplus W,\qquad W=Z\cap\ker R.
$$
If $P:E\to Z$ is a bounded projection, then $(I-R)P$ is a bounded projection onto $W$: its range lies in $W$, and it fixes every element of $W$. Hence $W$ is a <complemented subspace> of $E$, so $E\cong W\oplus V$ for some closed $V$.
We give the absorption argument in the <Pełczyński decomposition method> explicitly. Reindexing coordinates by a bijection between $\mathbb N^2$ and $\mathbb N$ yields $E\cong(\bigoplus_{n\ge1}E)_p$. Applying the fixed <Banach space isomorphism> $E\cong W\oplus V$ in every coordinate gives
$$
E\oplus W
\cong\left(\bigoplus_{n\ge1}(W\oplus V)\right)_p\oplus W
\cong\left(\bigoplus_{n\ge1}W\right)_p\oplus\left(\bigoplus_{n\ge1}V\right)_p\oplus W
\cong E.
$$
The extra $W$ is absorbed by shifting the countably many $W$ summands. The maps are bounded with bounded inverses because the same two-space decomposition is used uniformly in each summand; we may equip each finite <direct sum> with its equivalent $p$-sum norm. Since $Z\cong E\oplus W$, \b[$Z\cong\ell^p$].
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