[proofplan]
We prove both implications directly from the definitions. If $T$ preserves distances, then applying the distance-preserving identity to the pair $(x,0)$ and using linearity gives preservation of the norm of each vector. Conversely, if $T$ preserves the norm of every vector, then applying that hypothesis to the difference $x-y$ and using linearity shows that $T$ preserves the distance between arbitrary points.
[/proofplan]
[step:Derive norm preservation from the distance-preserving property]
Assume that $T:X\to Y$ is an isometry from $(X,d_X)$ to $(Y,d_Y)$, meaning that
\begin{align*}
d_Y(Tx,Ty)=d_X(x,y)
\end{align*}
for all $x,y\in X$.
Since $T$ is linear, $T0_X=0_Y$. Therefore, for every $x\in X$,
\begin{align*}
\|Tx\|_Y=d_Y(Tx,0_Y).
\end{align*}
Because $T0_X=0_Y$, this becomes
\begin{align*}
d_Y(Tx,0_Y)=d_Y(Tx,T0_X).
\end{align*}
Using the isometry property with the pair $(x,0_X)$ gives
\begin{align*}
d_Y(Tx,T0_X)=d_X(x,0_X).
\end{align*}
By the definition of $d_X$,
\begin{align*}
d_X(x,0_X)=\|x-0_X\|_X=\|x\|_X.
\end{align*}
Combining these equalities gives
\begin{align*}
\|Tx\|_Y=\|x\|_X
\end{align*}
for every $x\in X$.
[guided]
Assume that $T:X\to Y$ is an isometry from $(X,d_X)$ to $(Y,d_Y)$. By definition, this means that $T$ preserves the distance between every pair of points:
\begin{align*}
d_Y(Tx,Ty)=d_X(x,y)
\end{align*}
for all $x,y\in X$.
To recover a norm from a metric induced by a norm, we measure distance from the zero vector. Since $T$ is linear, it sends the zero vector of $X$ to the zero vector of $Y$:
\begin{align*}
T0_X=0_Y.
\end{align*}
Fix an arbitrary vector $x\in X$. By the definition of the metric $d_Y$,
\begin{align*}
\|Tx\|_Y=\|Tx-0_Y\|_Y=d_Y(Tx,0_Y).
\end{align*}
Using $T0_X=0_Y$, we may rewrite the second argument as $T0_X$:
\begin{align*}
d_Y(Tx,0_Y)=d_Y(Tx,T0_X).
\end{align*}
Now the isometry hypothesis applies to the pair $(x,0_X)$, so
\begin{align*}
d_Y(Tx,T0_X)=d_X(x,0_X).
\end{align*}
Finally, the definition of $d_X$ gives
\begin{align*}
d_X(x,0_X)=\|x-0_X\|_X=\|x\|_X.
\end{align*}
Putting these equalities together yields
\begin{align*}
\|Tx\|_Y=\|x\|_X.
\end{align*}
Since $x\in X$ was arbitrary, $T$ preserves the norm of every vector in $X$.
[/guided]
[/step]
[step:Derive the distance-preserving property from norm preservation]
Assume that
\begin{align*}
\|Tz\|_Y=\|z\|_X
\end{align*}
for every $z\in X$.
Let $x,y\in X$ be arbitrary. By the definition of $d_Y$,
\begin{align*}
d_Y(Tx,Ty)=\|Tx-Ty\|_Y.
\end{align*}
Since $T$ is linear,
\begin{align*}
Tx-Ty=T(x-y).
\end{align*}
Therefore
\begin{align*}
d_Y(Tx,Ty)=\|T(x-y)\|_Y.
\end{align*}
Applying the norm-preservation hypothesis to the vector $z=x-y\in X$ gives
\begin{align*}
\|T(x-y)\|_Y=\|x-y\|_X.
\end{align*}
By the definition of $d_X$,
\begin{align*}
\|x-y\|_X=d_X(x,y).
\end{align*}
Thus
\begin{align*}
d_Y(Tx,Ty)=d_X(x,y)
\end{align*}
for all $x,y\in X$. Hence $T$ is an isometry from $(X,d_X)$ to $(Y,d_Y)$.
[guided]
Assume that $T$ preserves the norm of every vector:
\begin{align*}
\|Tz\|_Y=\|z\|_X
\end{align*}
for all $z\in X$.
To prove that $T$ is an isometry, we must show that it preserves distances between arbitrary points. Let $x,y\in X$ be arbitrary. The distance between $Tx$ and $Ty$ in $Y$ is, by definition of the metric induced by $\|\cdot\|_Y$,
\begin{align*}
d_Y(Tx,Ty)=\|Tx-Ty\|_Y.
\end{align*}
The reason linearity matters is that the difference of the images is the image of the difference. Since $T$ is linear,
\begin{align*}
Tx-Ty=T(x)-T(y)=T(x-y).
\end{align*}
Substituting this into the distance formula gives
\begin{align*}
d_Y(Tx,Ty)=\|T(x-y)\|_Y.
\end{align*}
Now $x-y$ is a vector in $X$, so the norm-preservation hypothesis applies to the particular vector $z=x-y$. Hence
\begin{align*}
\|T(x-y)\|_Y=\|x-y\|_X.
\end{align*}
By the definition of the metric $d_X$ induced by $\|\cdot\|_X$,
\begin{align*}
\|x-y\|_X=d_X(x,y).
\end{align*}
Combining these equalities gives
\begin{align*}
d_Y(Tx,Ty)=d_X(x,y).
\end{align*}
Since the vectors $x,y\in X$ were arbitrary, $T$ preserves the distance between every pair of points in $X$. Therefore $T$ is an isometry from $(X,d_X)$ to $(Y,d_Y)$.
[/guided]
[/step]
