Real Analysis, Folland problem 5.5.55 Hilbert Spaces 
problem 3.5.55 - Let $\mathcal{H}$ be a Hilbert space.
a.) (The polarization identity) For any $x,y\in\mathcal{H}$ $$<x,y> = \frac{1}{4}(\lVert x + y \rVert^2 - \lVert x - y\rVert^2 + i\lVert x + iy\rVert^2 - i\lVert x - iy\rVert^2)$$
b.) If $\mathcal{H}'$ is another Hilbert space, a linear map from $\mathcal{H}$ to $\mathcal{H}'$ is unitary if and only if it is isometric and surjective.

Attempted proof a.)$$\lVert x + y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2 + <x,y> + <y,x> $$
$$-\lVert x - y \rVert^2 = -\lVert x \rVert^2 - \lVert y \rVert^2 + <x,y> + <y,x> $$
$$i\lVert x + iy\rVert^2 = i\lVert x \rVert^2 + i\lVert y \rVert^2 +  <x,y> - <y,x> $$
$$-i\lVert x - iy\rVert^2 = -i\lVert x \rVert^2 - i\lVert y \rVert^2 +  <x,y> - <y,x> $$
Adding the above gives the result
Attempted proof b.): Suppose, $f$ is a unitary linear map from $\mathcal{H}$ to $\mathcal{H}'$. We need to show that $f$ is isometric and surjective. Since $f$ is a unitary linear map we know from Folland that $f$ is also an invertible linear map that preserves inner products: $$\langle fx,fy\rangle_{2} = \langle x,y\rangle _{1} \ \ \forall \ x,y\in\mathcal{H}$$
Set $y = x$, then we have $\lVert fx\rVert_{2} = \lVert x\rVert_{1}$ thus an isometry.
I am not sure how to proceed further or if this approach is incorrect somehow. Any suggestions is greatly appreciated, I will update more as I keep thinking about it.
 A: You have proved that it is an isometry. It must be surjective for else it wouldn't be invertible.
Now assume that we have a surjective isometric linear mapping. We have to prove that it preserves inner products.
$<fx,fy>=\frac{1}{4}(||fx+fy||^2+||fx-fy||^2+i||fx+ify||^2-i||fx-ify||^2)$
$=\frac{1}{4}(||f(x+y)||^2+||f(x-y)||^2+i||f(x+iy)||^2-i||f(x-iy)||^2)$
$=\frac{1}{4}(||x+y||^2+||x-y||^2+i||x+iy||^2+||x-iy||^2)$
$=<x,y>$
QED.
A: Unitary means invertible and inner-product-preserving:
$$\langle fx,fy\rangle_2=\langle x,y\rangle_1.\tag1$$
On the other hand, surjectivity is logically "half" of bijectivity, and isometric means $$\lVert fx\rVert_2 = \lVert x\rVert_1.\tag2$$
However, a linear map $f:\mathcal{H}\to\mathcal{H}'$ is invertible if it's bijective and its inverse is bounded:
$$\lVert fx\rVert_2\ge c\lVert x\rVert_1
\qquad\text{for some}\quad c>0.\tag3$$
As you noted (and Folland does right after the definition of a unitary map, on page 176), setting $y=x$ shows that $(1)\implies(2)$, inner-product-preserving linear (unitary) maps are isometries. Since unitary maps are also invertible, they must be surjective. This proves one direction: an invertible inner-product-preserving linear map is a surjective isometry.
However (to prove that a surjective isometric linear map is invertible and inner-product-preserving), note that an isometry also has bounded inverse, $(2)\implies(3)$, as seen from their definitions on page 154, taking $c=1$. Isometries are clearly injective since $x\ne y\implies$ $0\ne\lVert{x-y}\rVert=$ $\lVert{f(x-y)}\rVert=$ $\lVert{f(x)-f(y)}\rVert\implies$ $f(x)\ne f(y)$, but if we are given that $f$ is also surjective, then it is therefore bijective and hence also invertible, by definition. We thus only need to show that a surjective isometry is also inner-product-preserving. For this, we'll use the polarization identity from part a to show that preserving the norm entails preserving the inner product.
To verify part a, first note what Folland uses in the proof of the Parallelogram Law, proposition 5.22:
$$\lVert x\pm y\rVert^2=\lVert x\rVert^2\pm2\,
\text{Re}\langle x,y\rangle+\lVert y\rVert^2,$$
where $2\text{Re}\langle{x,y}\rangle=
\langle{x,y}\rangle+\langle{y,x}\rangle$
since the summed terms are complex conjugates.
Noting that
$$\lVert x\pm iy\rVert^2=\lVert x\rVert^2\pm2\,
\text{Re}\langle x,iy\rangle+\lVert iy\rVert^2,$$
and $\langle x,iy\rangle=-i\langle x,y\rangle$ so that
$$\text{Re}\langle x,iy\rangle=\text{Im}\langle x,y\rangle,$$
we conclude that
$$\begin{align}
\langle x,y\rangle
&=\text{Re}\langle x,y\rangle+i\,\text{Im}\langle x,y\rangle\\
&=\text{Re}\langle x,y\rangle+i\,\text{Re}\langle x,iy\rangle\\
&=\tfrac14\left[
\left(\lVert x+y\rVert^2-\lVert x-y\rVert^2\right)+i
\left(\lVert x+iy\rVert^2-\lVert x-iy\rVert^2\right)\right]\\
&=\tfrac14\left(
\lVert x+y\rVert^2-\lVert x-y\rVert^2+i
\lVert x+iy\rVert^2-i\lVert x-iy\rVert^2
\right).
\end{align}$$
Note the negative sign by the second term, correcting a typo in the book for part a. Finally, the conclusion of part b follows from $(2)$ since $f$ is an isometry:
$$\begin{align}
\langle fx,fy\rangle
&=\tfrac14\left(
\lVert fx+fy\rVert^2-\lVert fx-fy\rVert^2+i
\lVert fx+ify\rVert^2-i\lVert fx-ify\rVert^2
\right)\\
&=\tfrac14\left(
\lVert f(x+y)\rVert^2-\lVert f(x-y)\rVert^2+i
\lVert f(x+iy)\rVert^2-i\lVert f(x-iy)\rVert^2
\right)\\
&=\tfrac14\left(
\lVert x+y\rVert^2-\lVert x-y\rVert^2+i
\lVert x+iy\rVert^2-i\lVert x-iy\rVert^2
\right)\\
&=\langle x,y\rangle.
\end{align}$$
