Squeeze theorem for duals of Banach spaces

Suppose that $X\subseteq Y \subseteq Z$ are Banach spaces such that $X$ is complemented in $Y$ and the duals $X^*, Z^*$ are isomorphic. Must the dual $Y^*$ be isomorphic to $X^*$?

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I don't think so. Here's how I would go about constructing an example. Start with $E$ such that $E \times E \cong E$. Choose $F \lt E$ non-complemented. Put $X = E$, $Y = E \times F$ and $Z = E \times E$. –  t.b. Sep 2 '11 at 0:23
Thank you. However in the particular case I am interested in, $Z$ is not isomorphic to its square. –  Robert Mateja Sep 2 '11 at 0:32

The answer is no, even if you assume that Z does not contain an isomorphic copy of its square. For instance, take $X=c_0(\omega_1)$, $Y= c_0(\omega_1)\oplus E$, where $E$ is a subspace of $c_0$ such that $E^*$ is not isomorphic to $\ell_1$, and $Z=C([0,\, \omega_1])$. Since the $c_0$-direct sum of any sequence of finite dimensional Banach spaces is almost-isometric to a subspace of $c_0$, you could take $E=(\bigoplus_{n=1}^\infty \ell_1^n)_{c_0}$. To see that $E^*$ is not isomorphic to $\ell_1$, notice that $c_0$ is not crudely finitely representable in $\ell_1$, whereas $c_0$ is crudely finitely representable in $(\bigoplus_n \ell_1^n)_{c_0}^*$ since $(\bigoplus_n \ell_1^n)_{c_0}^*$ contains uniform copies of $\ell_\infty^n$.