Let $X,Y$ be a normed spaces, not necessarily finitely dimensional and let $T\in B(X,Y)$ be an isometric isomorphism. I want to show the same holds for the dual of $T$, $T'\in B(Y',X')$. The isomorphism part is easy, since one can easily show that $T'$ is invertible whenever $T$ is. Now, I want to show that $T'$ is an isometry. Is the following correct?
$ \|T'(y')\|=\sup\{|y'(Tx)|\,\big|\,\|x\|\leq1\}=\sup\{|y'(Tx)|\,\big| \|Tx\|\leq 1\}=\|y'\| $
I am not sure about the part where I substituted $\|Tx\|$ for $\|x\|$, but I think it is allowed since $T$ is an isometry.
Yes, it is correct. As $T$ is an isometry, $\|Tx\|=\|x\|$ for all $x\in X$. Hence we have an equality of sets $$\{|y'(Tx)|:\|x\|\leq1\}=\{|y'(Tx)|:\|Tx\|\leq1\},$$ and taking suprema yields the desired equality.