Let $T:X\rightarrow X$ be an operator on a Hilbert space $X$ and assume it is unitary, i.e. $T^*T=TT^*=I$. I want to show that $\|T\|=1$
We know that for any operator $\|TT^*\|=\|T\|^2$ but since $T$ is unitary we also know that $\|TT^*\|=\|I\|=\sup_{\|x\|=1}\|I(x)\|=\sup_{\|x\|=1}\|x\|=1$ therefore if we combine both we get $$\|T\|^2=1\Leftrightarrow \|T\|=1$$does this work?
It depends on the vector norm $\|x\|$ you use. If it is the norm induced by the inner product with respect to which $T^\ast$ is defined, then $\|Tx\|^2=\langle Tx,Tx\rangle^2=\langle x,T^\ast Tx\rangle^2=\|x\|^2$ for every vector $x$ and hence $\|T\|=1$.
However, if $\|x\|$ is some norm other than $\sqrt{\langle x,x\rangle}$, it can happen that $\|T\|>1$. For instance, when $$ T=\pmatrix{\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}}, $$ we have $\|T\|_\infty=\sup_{\|x\|_\infty=1}\|Tx\|_\infty=\sqrt{2}$.