Every $\ell^p$ space with $p\ne 2$ has a subspace without a Schauder basis

Question

Rajendra Bhatia's Notes on Functional Analysis (Texts and Readings in Mathematics), Pg. $14$ states (without proof) that:

Every $\ell^p$ space ($1\le p \le \infty)$ with $p\ne 2$ has a subspace without a Schauder basis.

where $\ell^p$ denotes the sequence space with the $p$-norm.

1. What is the proof of this fact? I have been trying for a while now, and it seems more difficult than I had imagined.

2. Also, I would be interested to see why every subspace of $\ell^2$ has a Schauder basis.

P.S. If the proof is doable with some hints, then just hints would be great too!

Update: The proof of the main assertion is found in Lindenstrauss and Tzafriri's Classical Banach Spaces I and II, as mentioned by David Mitra in the comments.

Answer 1

The answer to question (1) is in the comment by @DavidMitra: You can refer to Lindenstrauss' and Tzafriri's Classical Banach Spaces I and II for a proof that every $\ell^p$ space with $p \neq 2$ admits a subspace without a Schauder basis.

The answer to question (2) is far easier: note that every subspace of $\ell^2$ is a separable inner product space. Since every separable inner product space admits a (countable) orthonormal basis (e.g. @mechanodroid answer here), you can take this orthonormal basis as your Schauder basis.