Let $\displaystyle V$ be an inner product space, $\displaystyle T:V \rightarrow V$ be a linear operator such that $\displaystyle ||T(x)||=||x||$ show that $\displaystyle T$ is injective.
No. While it is true that if $\displaystyle \left(V,\langle\cdot,\cdot\rangle\right)$ is an inner product space there is a canonical way to make it into a normed vector space $\displaystyle \left(V,\|\cdot\|\right)$ by defining $\displaystyle \|v\|=\sqrt{\langle v,v\rangle}$. The opposite is not necessarily true. There is no (at least not that I know of ) natural way of defining $\displaystyle \langle \cdot,\cdot\rangle$ given $\displaystyle \|\cdot\|$.
The obvious reason is that to go $\displaystyle \langle \cdot,\cdot\rangle\to\|\cdot\|$ notice we were reducing a bivalued function to a single valued function. There is an obvious way to do this by replacing both arguments of the bivalued function by one value. To go the other way is not so easy.
So, my point is that in general $\displaystyle \text{Inner product spaces}\subset\text{Normed vector spaces}$ and we only need the latter opposed to the former, so it's strange they gave us an "extra" condition.
For fun, see if you can find an easy inner product using the norm $\displaystyle \|(x_1,\cdots,x_n)\|=\max_{1\leqslant j\leqslant n}|x_j|$ with $\displaystyle (x_1,\cdots,x_n)\subseteq\mathbb{R}^n$
EDIT: An easy way to think about this is $\displaystyle \langle \cdot,\cdot\rangle\implies\text{ A notion of angles}$ and $\displaystyle \|\cdot\|\implies\text{ A notion of length}$. One clearly follows from the other but not vice versa.