> If $T\colon X\to Y$ is isometric isomorphism, then so is $T^*\colon Y^* \to X^*$.

Unless I am missing something, this should be obvious or at most routine verification. It basically says that if two space are "the same" then the duals are "the same", too.

Yet I see in the proof of Theorem 1.10.12 in Megginson's Banach Space Theory that the author uses, for example, a consequence of Open Mapping Theorem (Corollary 1.6.6).

Am I missing something which is not so straightforward?

I suppose if we only wanted to prove this last part of the theorem (and not also other things mentioned there), this part would be immediate, right?