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$\mathbb{R}^n\not\cong\mathbb{R}^n\setminus\{0\}$ are not homeomorphic is a triviality from Algebraic Topology. On the other hand, if $X$ is an infinite dimensional Banach space, then $X \cong X\...

$B(\ell^2)$ is an injective operator system by a result of Arveson. However, $B(\ell^2)$ is not an injective Banach space, since it is not linearly isomorphic to a $C(K)$ space (for instance, $C(K)$ ...