Is this alternative argument valid? 
Show that the sup norm on $\mathbb{R}^2$ is not derived from an inner product $\mathbb{R}^2$. [Hint: Suppose $\langle x,y \rangle$ (not the dot product) having the property that $|x| = \langle x,x \rangle^{1/2}$. Compute $\langle x\pm y, x\pm y \rangle$ and apply it to the case $\mathbf{x} = \mathbf{e_1}$ and $\mathbf{y} = \mathbf{e_2}$]

The book defines the sup-norm as $$|x| = \max \{|x_1|,|x_2|,\dots\}$$
So I thought about just comparing $|e_1 \pm e_2| = \max\{|1|,|\pm1|\} = 1$ by sup-norm and $\langle e_1 \pm e_2, e_1 \pm e_2 \rangle^{1/2} = |e_1 \pm e_2| = \sqrt{2}$ and this shows they are differently, so the supnorm could not possibly have any relationship with the inner product?
I am also a little confused by the purpose of the question, why do we have to show that sup norm isn't derived from the inner product? Isn't that obvious? Why would they have any relationship in the first place?
edit to reply deleted answer How is that [parallelogram law method], in theory, different from what I did? In the parallelogram law method, we just looked at the maximum of both sides and checked we get 2 = 4. I didn't bother with that and got 1 = $\sqrt{2}$. 
 A: The dot product is only one such inner product. Many other inner products can be defined.
The statement that
$$\langle e_1 \pm e_2, e_1 \pm e_2 \rangle^{1/2} = \sqrt{2}$$ is true only if the inner product you are talking about is the dot-product.In general, it need not be $\sqrt{2}$ for other inner products.
The goal of the question is to show that no inner product (not necessarily the dot product) can induce the sup-norm.
To check this, any norm obtained from the inner-product should satisfy the "parallelogram law". Whereas the $p$-norm with $p \neq 2$, (sup-norm corresponds to the case $p \to \infty$) does not satisfy the parallelogram law.
And conversely, any norm satisfying the parallelogram law comes from an inner product. This is a nice exercise and is often called the "polarisation identity".
A: Just as an aside: there is a simpler^W more conceptual way of proving this. All inner product $\mathbb{R}$-vector spaces of a finite dimension $n$ are isomorphic because they all look like the Euclidean inner product in their orthonormal bases (which exist by Gram-Schmidt).
Note that this means that their unit spheres are then the Euclidean unit sphere transformed by the corresponding linear isomorphisms. Since these isomorphisms are also diffeomorphisms, a unit sphere of any norm derived from an inner product must be a smooth embedding of $S^{n-1}$, which is clearly not the case for the supremum norm.
Alternatively, instead of using differential geometry, we can use algebraic geometry (which I'm less familiar with, so I may make mistakes below). The linear isomorphisms discussed earlier are also biregular, so the unit sphere of any norm derived from an inner product must be an algebraic variety biregular to the Euclidean unit sphere, which is irreducible. But the unit sphere of the supremum norm is not Zariski closed, in fact its Zariski closure is a union of $2n$ hyperplanes, and is thus reducible.
