Show that if $X$ is a Hilbert space, then so is $Y$ Given that $X$ and $Y$ are normed linear spaces and that $T:X\to Y$ is a linear map, such that $T(\alpha x_1 + \beta x_2) = \alpha T(x_1)+ \beta T(x_2)$ for all vectors $x_1,x_2\in X$ and scalars $\alpha , \beta$. Suppose $T$ maps $X$ onto $Y$ and is isometric, $\|Tx\|=\|x\|$ for all $x\in X$.
1) Show that if $X$ is a Hilbert space then so is $Y$ if we define:
$$\langle y_1,y_2\rangle_Y=\langle x_1,x_2\rangle_X$$
Where $x_1,x_2$ are the unique points in $X$ satisfying $Tx_1=y_1$ and $Tx_2=y_2$
What I have done so far
I have previously proved that if $X$ is a Banach space then so is $Y$.
So I am thinking that since $X$ is a Hilbert space, 
$$||y_1-y_2|| = ||T(x_1)-T(x_2)|| = ||T(x_1-x_2)||=||x_1-x_2|| \implies$$
$$||y_1-y_2||^2 = ||y_1||^2 - 2Re\langle y_1,y_2 \rangle + ||y_2||^2$$
$$||x_1-x_2||^2 = ||x_1||^2 - 2Re\langle x_1,x_2 \rangle + ||x_2||^2$$
and
$$||y_1+y_2|| = ||T(x_1)+T(x_2)|| = ||T(x_1+x_2)||=||x_1+x_2|| \implies$$
$$||y_1+y_2||^2 = ||y_1||^2 + 2Re\langle y_1,y_2 \rangle + ||y_2||^2$$
$$||x_1+x_2||^2 = ||x_1||^2 + 2Re\langle x_1,x_2 \rangle + ||x_2||^2$$
Therefore
$$||y_1+y_2||^2+||y_1-y_2||^2=2||y_1||^2+2||y_2||^2$$
Therefore the parallelogram equality holds for $Y$, and therefore there is an inner product that gives the norm, and so $||y_1 - y_2||=\sqrt{\langle y_1-y_2,y_1-y_2 \rangle}$ and therefore $Y$ is a Hilbert space.

You should go this way

$$ ||y_1-y_2|| = ||T(x_1)-T(x_2)|| = ||T(x_1-x_2)||=||x_1-x_2||$$

since you have been given

$$ ||T(x)||=||x|| $$

and hence the parallelogram law will hold.
