Prove that an Isometric linear operator on a finite dimensional inner product space is Unitary I really don't find relevance of being finite dimensional of the inner product in proving an isometric linear operator is unitary.Please help me 

If $T$ is an isometry, then
$$\langle T^*Tx,y \rangle = \langle Tx,Ty\rangle=\langle x,y\rangle$$
holds for every $x,y$. Hence $T^*T=I$. Since the space is finite dimensional, we must also have $TT^*=I$. Therefore $T$ is unitary.

This isn't true in infinite dimensions. Indeed, on the Hilbert space
$$\ell^2 := \{x:\mathbb{N}\to\mathbb{C} \mid \sum_{n=1}^\infty |x(n)|^2 < \infty\}$$
with inner product
$$ \langle x,y \rangle = \sum_{n=1}^\infty x(n)\overline{y(n)},
$$
the "right shift operator" $R:\ell^2\to\ell^2$ defined by
$$R(x)(n)=\begin{cases} 0 & \text{if}\ n=1 \\ x(n-1) & \text{otherwise} \end{cases}$$
is an isometry that is not even a surjection, and thus not unitary.
