Q:

Is linear surjective isometry always unitary? 

Basically what I'm trying to show is $\forall h_1, \ h_2 \in \mathscr{H}$ and $U: \mathscr{H} \rightarrow \mathscr{K}$ then $\langle Uh_1, \ Uh_2\rangle_\mathscr{K} = \langle h_1, \ h_2 \rangle_\mathscr{H}$  (So your standard definition).  Now I figure the easiest way to do so is by polarization. 
Since we have an isometry then 
$\begin{eqnarray*}
\langle h_1, \ h_2 \rangle_\mathscr{H} &=& \frac{1}{4} \bigg( \|h_1 + h_2\|_\mathscr{H}- \|h_1-h_2\|_\mathscr{H} +i\|h_1+ih_2\|_\mathscr{H} - i\|h_1 - ih_2\|_\mathscr{H} \bigg)
\\&=&\frac{1}{4} \bigg( \|Uh_1 + Uh_2\|_\mathscr{K}- \|Uh_1-Uh_2\|_\mathscr{K} +i\|Uh_1+iUh_2\|_\mathscr{K} - i\|Uh_1 - Uih_2\|_\mathscr{K} \bigg)
\\&=& \langle Uh_1, \ Uh_2\rangle_\mathscr{K}
\end{eqnarray*} $
But what I'm wondering is where surjectivity comes into play? 

A:

Let $U: \mathcal{H} \rightarrow \mathcal{H}$ be an isometric and surjective operator. 
We want to show that $U^* U = U U^* = I,$ i.e. the adjoint of $U$ is its inverse. 
You have shown that isometry implies $\langle U h_1 , U h_2 \rangle =\langle  h_1 , h_2 \rangle$ for all $h_1,h_2 \in \mathcal{H}$. By 
$$\langle  h_1 , h_2 \rangle = \langle U h_1 , U h_2 \rangle = \langle U^*U h_1 , h_2 \rangle, $$This implies $U^* U = I,$ i.e. $U^*$ is a left-inverse of $U$, which implies that $U$ is injective. 
Together with the surjectivity, this implies that $U$ is indeed bijective and the left-inverse coincides with the proper inverse. 

A:

A unitary operator should also be invertible.  A linear isometry is merely injective, not necessarily surjective, so you need to take on that requirement.  In other words, the condition you have that $\langle Uh_1, Uh_2 \rangle = \langle h_1, h_2 \rangle$ is necessary but not sufficient for unitarity.  You also want $\langle U^* h_1, U^* h_2 \rangle = \langle h_1, h_2 \rangle$.

