Are all Banach spaces also Hilbert spaces? We have the well-known "polarization identity"
$$(x,y)=\frac{1}{4}\left(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2\right)\tag{1}$$
that works in any Hilbert space. Hence, is every Banach space also a Hilbert space by just defining the inner product $(\cdot,\cdot)$ by $(1)$?
The claim is clearly false, but I cannot find why.

A Banach space is a complete metric space with a norm, while a Hilbert space is a complete metric space with an inner product. An inner product induces a norm by $||x|| = \sqrt{<x,x>}$, which means that any Hilbert space is also a Banach space. However, the converse need not hold as the norm isn't always expressable in terms of the inner product.
An example is $C(K, \mathbb{R})$, the space of all continuous functions on a compact  set $K$ into $\mathbb{R}$ with the norm $||f|| = \sup_{x \in K}{|f(x)|}$. 
