Currently I am reading Topics in Banach space Theory by Albiac and Kalton, $2$nd edition.
The authors provided a proof of Pitt's theorem at page $31,$ chapter $2.$
Theorem $2.1.4$ (Pitt's Theorem) Suppose $1\leq p<r <\infty.$ If $X$ is a closed subspace of $\ell_r$ and $T:X\to \ell_p$ is a bounded linear operator, then $T$ is compact.
Straight after the proof of Pitt's theorem, the authors provided the following corollary (in between the theorem and corollary, there is a remark on the proof of Pitt's theorem, thus the numbering).
Corollary $2.1.6$ The spaces of the set $\{c_0\}\cup \{\ell_p:1\leq p<\infty\}$ are mutually non isomorphic. In fact, if $X$ is an infinite-dimensional subspace of one of the spaces $\{c_0\}\cup \{\ell_p:1\leq p<\infty\}$, then $X$ is not isomorphic to a subspace of any other.
The authors does not provide any proof to the corollary. I have trouble identifying the corollary is due to Pitt's theorem.
The following is my attempted proof to the corollary.
Assume that there is an isomorphism $T:X\to \ell^p$ where $X\subseteq \ell^q$ is a closed subspace. Since $T$ and $T^{-1}$ are both bounded, without loss of generality, we can assume that $1\leq p<q<\infty.$ By Pitt's theorem, $T$ is a a compact operator. Note that $$B_{\ell^p}\subseteq T(\|T^{-1}\| B_{X})$$ where $B_{\ell^p}$ and $B_X$ are the closed unit ball of $\ell^p$ and $X$ respectively. Since $T$ is compact, by definition of compact operator, $\overline{T(\|T^{-1}\| B_{X})}$ is compact. Since $T(\|T^{-1}\| B_{X})$ is closed, therefore $T(\|T^{-1}\| B_{X})$ is compact. Since $B_{\ell^p}$ is a closed subset of a compact set $T(\|T^{-1}\| B_{X}),$ we have $B_{\ell^p}$ is compact as well. This implies that $\ell^p$ is finite-dimensional, contradiction.
Is my proof above correct?