Prove $\|T(x)\| = \|x\|$ for all $x \in V$ if and only if $\langle T(x),T(y)\rangle = \langle x,y\rangle$ for all $x,y \in V$. Let $T$ be a linear operator on inner product space $V$. Prove $\|T(x)\| = \|x\|$ for all $x \in V$ if and only if $\langle T(x),T(y)\rangle = \langle x,y\rangle$ for all $x,y \in V$.
For the reverse direction, it is easy, we just set x=y, what to do with the forward direction?
What facts to use to prove this and general idea in approaching this problem?
 A: In essence, your proof utilized the fact that you can write the norm in terms of the inner product. To do the reverse direction, you'd want to write the inner product in terms of the norm. I hope this is enough to start you off. I'll include a solution as a spoiler below.

 We compute that $|x + y|^2 = \langle x + y, x + y \rangle = \langle x, x + y \rangle + \langle y, x + y \rangle = \langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle$. Rewriting this, we get $|x + y|^2 = |x|^2 + 2 \langle x, y \rangle + |y|^2$. Hence, $\langle x, y \rangle = \frac{1}{2} (|x + y|^2 - |x|^2 - |y|^2)$. Hence, if $|T(x)| = |x|$ for all $x$ then $\langle T(x), T(y) \rangle = \frac{1}{2} (|T(x) + T(y)|^2 - |T(x)|^2 - |T(y)|^2) = \frac{1}{2} (|x + y|^2 - |x|^2 - |y|^2) = \langle x, y \rangle.$

A: You can approach it alternatively as
\begin{align*}
\langle T(x+y),T(x+y)\rangle & = \langle T(x) + T(y),T(x) + T(y)\rangle\\\\
& = \langle T(x),T(x)\rangle + 2\langle T(x),T(y)\rangle + \langle T(y),T(y)\rangle\\\\
& = \|T(x)\|^{2} + 2\langle T(x),T(y)\rangle + \|T(y)\|^{2}
\end{align*}
On the other hand, one has that
\begin{align*}
\langle x+y,x+y\rangle & = \langle x,x\rangle + 2\langle x,y\rangle + \langle y,y\rangle\\\\
& = \|x\|^{2} + 2\langle x,y\rangle + \|y\|^{2}
\end{align*}
Since both expressions are equal and $T$ is an isometry, we can conclude that
\begin{align*}
\|T(x)\|^{2} + 2\langle T(x),T(y)\rangle + \|T(y)\|^{2} = \|x\|^{2} + 2\langle x,y\rangle + \|y\|^{2} \Longrightarrow \langle T(x),T(y)\rangle = \langle x,y\rangle
\end{align*}
and we are done with case where the field $\textbf{F} = \textbf{R}$.
Let us suppose now that $\textbf{F} = \textbf{C}$. To begin with, we should notice that
\begin{align*}
\langle T(x+y),T(x+y)\rangle & = \langle T(x) + T(y),T(x) + T(y)\rangle\\\\
& = \langle T(x),T(x)\rangle + \langle T(x),T(y)\rangle + \langle T(y),T(x)\rangle + \langle T(y),T(y)\rangle\\\\
& = \|T(x)\|^{2}  + \langle T(x),T(y)\rangle + \overline{\langle T(x),T(y)\rangle} + \|T(y)\|^{2}\\\\
& = \|T(x)\|^{2} + 2\text{Re}\langle T(x),T(y)\rangle + \|T(y)\|^{2}
\end{align*}
Similarly, one does also have that
\begin{align*}
\langle x+y,x+y\rangle & = \langle x,x\rangle + \langle x,y\rangle + \langle y,x\rangle + \langle y,y\rangle\\\\
& = \|x\|^{2} + \langle x,y\rangle + \overline{\langle x,y\rangle} + \|y\|^{2}\\\\
& = \|x\|^{2} + 2\text{Re}\langle x,y\rangle + \|y\|^{2}
\end{align*}
Therefore, since both expressions are equal and $T$ is an isometry, $\text{Re}\langle T(x),T(y)\rangle = \text{Re}\langle x,y\rangle$.
On the other hand, one has that
\begin{align*}
\langle T(x+iy),T(x+iy)\rangle & = \langle T(x) + iT(y), T(x) + iT(y)\rangle\\\\
& = \langle T(x),T(x)\rangle + \langle T(x),iT(y)\rangle + \langle iT(y),T(x)\rangle + \langle iT(y),iT(y)\rangle\\\\
& = \|T(x)\|^{2} - i\langle T(x),T(y)\rangle + i\langle T(y),T(x)\rangle + \|T(y)\|^{2}\\\\
& = \|T(x)\|^{2} - i\langle T(x),T(y)\rangle + i \overline{\langle T(x),T(y)\rangle} + \|T(y)\|^{2}\\\\
& = \|T(x)\|^{2} + 2\text{Im}\langle T(x),T(y)\rangle + \|T(y)\|^{2}
\end{align*}
which is equal to
\begin{align*}
\langle x + iy,x+iy\rangle & = \langle x,x\rangle + \langle x,iy\rangle + \langle iy,x\rangle + \langle iy,iy\rangle\\\\
& = \|x\|^{2} - i\langle x,y\rangle + i\overline{\langle x,y\rangle} + \|y\|^{2}\\\\
& = \|x\|^{2} + 2\text{Im}\langle x,y\rangle + \|y\|^{2}
\end{align*}
Since $T$ is an isometry, we conclude that $\text{Im}\langle T(x),T(y)\rangle = \text{Im}\langle x,y\rangle$.
Finally, gathering both results, we conclude that $\langle T(x),T(y)\rangle = \langle x,y\rangle$, just as desired.
Hopefully this helps.
