The Schroeder-Bernstein problem asked about complemented subspaces. It is a much stronger property for a subspace to be complemented. The only spaces in which every subspace is complemented are isomorphic to Hilbert space (Lindenstrauss and Pelczynski). Also, Hilbert space (and its isomorphs) is the only space isomorphic to all of its subspaces. Therefore, if you consider any space which is not isomorphic to Hilbert space but embedds into its subspaces (e.g. $\ell_p $ for $p \not= 2$; in general, these are called minimal spaces), any non-isomorphic subspace gives a counterexample.

Edit: I guess I was confused by the term left invertible. I'll leave my answer up in case someone doesn't know how do do it in this weaker case.

The Schroeder-Bernstein problem asked about complemented subspaces. It is a much stronger property for a subspace to be complemented. The only spaces in which every subspace is complemented are isomorphic to Hilbert space (Lindenstrauss and Pelczynski). Also, Hilbert space (and its isomorphs) is the only space isomorphic to all of its subspaces. Therefore, if you consider any space which is not isomorphic to Hilbert space but embedds into its subspaces (e.g. $\ell_p $ for $p \not= 2$; in general, these are called minimal spaces), any non-isomorphic subspace gives a counterexample.

Edit: I guess I was confused by the term left invertible. I'll leave my answer up in case someone doesn't know how to do it in this weaker case.

The Schroeder-Bernstein problem asked about complemented subspaces. It is much strong property for a subspace to be complemented. The only spaces in which every subspace is complemented are isomorphic to Hilbert space (Lindenstrauss and Pelczynski). Also, Hilbert space (and its isomorphs) is the only space isomorphic to all of its subspaces. Therefore, if you consider any space which is not isomorphic to Hilbert space but embedds into its subspaces (e.g. $\ell_p $ for $p \not= 2$; in general, these are called minimal spaces), any non-isomorphic subspace gives a counterexample.

The Schroeder-Bernstein problem asked about complemented subspaces. It is a much stronger property for a subspace to be complemented. The only spaces in which every subspace is complemented are isomorphic to Hilbert space (Lindenstrauss and Pelczynski). Also, Hilbert space (and its isomorphs) is the only space isomorphic to all of its subspaces. Therefore, if you consider any space which is not isomorphic to Hilbert space but embedds into its subspaces (e.g. $\ell_p $ for $p \not= 2$; in general, these are called minimal spaces), any non-isomorphic subspace gives a counterexample.

Edit: I guess I was confused by the term left invertible. I'll leave my answer up in case someone doesn't know how do do it in this weaker case.