Most of linear algebra books define the norm for an inner product space $(V,\mathbb{F})$ as:
$$\|v\| = \sqrt{\langle v,v\rangle}$$
In which $v\in V$. From this definition, one can then state the following theorem:
Theorem. Let $V$ be an inner product space over $\mathbb{F}$. Then for all $x,y \in V$ and $c\in \mathbb{F}$, the following statements are true.
(a) $\|cx\| = |c|\cdot \|x\|$.
(b) $\|x\| = 0 \Longleftrightarrow x =0$. In any case, $\|x\| \geq 0$.
(c) $|\langle x,y\rangle | \leq \|x\| \cdot \|y\|$.
(d) $\|x+y\| \leq \|x\|+\|y\|$.
Problem is that in normed vector spaces, the norm does not necessarily comes from an inner product. One can verify that if the parallelogram identity:
$$\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2$$
Holds for a normed vector space, then $\|\cdot\|$ can be obtained from an inner product.
But then the definition for norm in a normed vector space is different from the definition given for inner product spaces. We define the norm for a normed vector space as items (a),(b) and (d) from the previous theorem.
Questions:
(1) Is the norm from a normed vector space not the 'same' as the norm from an inner product space?
(2) Why don't we define the norm in inner product spaces as we do in normed vector spaces? The norm definition for normed vector space seems more general since one could define a norm that does not necessarily come from an inner product. And even if it did come from an inner product, we could easily recover it via the parallelogram identity. Why would we want to restrict ourselves from using norms that do not come from an inner product?
Thanks everyone for all your comments, I think I can now answer my own question.
(1) The norm of a inner product space is indeed a norm. It's just a very special case in which the norm is induced by the inner product acting on the space.
(2) This was mostly a misconception due to textbooks defining $\lVert \cdot \rVert = \sqrt{\langle \cdot,\cdot\rangle}$ as the norm, rather than the norm induced by the inner product. So the definition for inner product spaces isn't defining all norms in general, but only those that come from the inner product. As @Mark and @Gonçalo mentioned, the theorem stated is proving the fact that the norm induced by the inner product is, indeed, a norm.
@BenGrossman also added extra information for this question: 'It's not that the definition of a norm over an inner product space is different from the more general definition of a norm, it's that every inner product space has a specific norm that is induced by and compatible with its inner product. It is certainly possible to define a norm over an inner product space that has nothing to do with the inner product, but the "natural" notion of a length in an inner product space comes from the function $v \mapsto \sqrt{\langle v,v\rangle}$ , which as the theorem you point to states is indeed a norm.'