Operator is an isometry iff |det| is 1

Question

$V$ is a finite dimensional real inner product space where $T\in \mathcal{L}(V)$ such that $\|Tv\|\leq\|v\|$ $\forall v\in V$.
Prove that $\left|\det T\right|=1 \iff T$ is an isometry.

My question is not for a proof of this but whether the following argument works.
I can use the theorem that $T$ is an isometry iff there exists an orthonormal basis of $V$ such that its matrix is block-diagonal with each block either of the form $\begin{bmatrix}\cos\theta & -\sin\theta \\\sin\theta & \cos\theta \end{bmatrix}$ or a $1\times 1$ block of either $-1$ or $1$. Then determinant of this diagonal matrix is must be $\pm 1$, and so $\left|\det T\right|=1$.
I believe this should prove the question for a generic operator $T\in \mathcal{L}(V)$ so we do not have to use the fact that $\|Tv\|\leq\|v\|$ $\forall v\in V$ and the answer should still be correct.
I know there are ways to solve this using $\|Tv\|\leq\|v\|$ but am specifically wondering if my reasoning above works, which I believe it should since if it is true for any real operator, then it is true with this restriction.