Solution 1:

Using the linearity of $T$, you have

$$ \left< T\alpha|T\beta \right>_W = \frac{1}{4} \left( ||T\alpha + T\beta||_W^2 - ||T\alpha - T\beta||_W^2 \right) = \frac{1}{4} \left( ||T(\alpha + \beta)||_W^2 - ||T(\alpha - \beta)||_W^2 \right) = \frac{1}{4} \left( ||\alpha + \beta||_V^2 - ||\alpha - \beta||_V^2 \right) = \left< \alpha | \beta \right>_V.$$